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<p>刚接触Flutter开发时，不管是为了性能还是项目代码简洁等原因，总归还是会选择一个合适的状态管理工具。但是看着茫茫多的选择还是感觉无从下手，当然我也一样。下方记录一下各种状态管理（Riverpod、Provider、Getx）的使用，以便后续遇到其他项目使用不同状态管理时能快速上手。也希望为看到这篇文章的同学提供一些参考。</p>
<h2 id="Riverpod"><a href="#Riverpod" class="headerlink" title="Riverpod"></a>Riverpod</h2><p>使用的前提需要注意，需要使用ProviderScope包裹父级节点。</p>
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<h3 id="1-二叉树的前序遍历"><a href="#1-二叉树的前序遍历" class="headerlink" title="1.二叉树的前序遍历"></a>1.<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/binary-tree-preorder-traversal/">二叉树的前序遍历</a></h3><p>给你二叉树的根节点 <code>root</code> ，返回它节点值的 <strong>前序</strong> 遍历。</p>
<p><strong>示例 1：</strong></p>
<p><img src="https://cdn.jsdelivr.net/gh/Tate-zwt/ImageHost/blog/202201071456889.jpg" alt="img"></p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：root = [1,null,2,3]</span><br><span class="line">输出：[1,2,3]</span><br></pre></td></tr></table></figure>
<p><strong>示例 2：</strong></p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：root = []</span><br><span class="line">输出：[]</span><br></pre></td></tr></table></figure>
<figure class="highlight swift"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Definition for a binary tree node.</span></span><br><span class="line"><span class="comment"> * public class TreeNode {</span></span><br><span class="line"><span class="comment"> * public var val: Int</span></span><br><span class="line"><span class="comment"> * public var left: TreeNode?</span></span><br><span class="line"><span class="comment"> * public var right: TreeNode?</span></span><br><span class="line"><span class="comment"> * public init() { self.val = 0; self.left = nil; self.right = nil; }</span></span><br><span class="line"><span class="comment"> * public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }</span></span><br><span class="line"><span class="comment"> * public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {</span></span><br><span class="line"><span class="comment"> * self.val = val</span></span><br><span class="line"><span class="comment"> * self.left = left</span></span><br><span class="line"><span class="comment"> * self.right = right</span></span><br><span class="line"><span class="comment"> * }</span></span><br><span class="line"><span class="comment"> * }</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>{</span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">preorderTraversal</span>(<span class="keyword">_</span> <span class="params">root</span>: <span class="type">TreeNode</span>?)</span> -> [<span class="type">Int</span>] {</span><br><span class="line"> <span class="keyword">var</span> result <span class="operator">=</span> [<span class="type">Int</span>]()</span><br><span class="line"> <span class="keyword">guard</span> <span class="keyword">let</span> root <span class="operator">=</span> root <span class="keyword">else</span> {</span><br><span class="line"> <span class="keyword">return</span> result</span><br><span class="line"> }</span><br><span class="line"> result.append(root.val)</span><br><span class="line"> result <span class="operator">+=</span> preorderTraversal(root.left)</span><br><span class="line"> result <span class="operator">+=</span> preorderTraversal(root.right)</span><br><span class="line"> <span class="keyword">return</span> result</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>解题思路：</p>
<ol>
<li>前序遍历按照根节点->左子树->右子树的顺序访问</li>
<li>访问根节点</li>
<li>采用中序递归遍历左子树</li>
<li>采用中序递归遍历右子树</li>
</ol>
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<h3 id="1-字符串轮转"><a href="#1-字符串轮转" class="headerlink" title="1.字符串轮转"></a>1.<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/string-rotation-lcci/">字符串轮转</a></h3><p>字符串轮转。给定两个字符串s1和s2，请编写代码检查s2是否为s1旋转而成（比如，waterbottle是erbottlewat旋转后的字符串）。</p>
<p>示例1:</p>
<p> 输入：s1 = “waterbottle”, s2 = “erbottlewat”<br> 输出：True<br>示例2:</p>
<p> 输入：s1 = “aa”, s2 = “aba”<br> 输出：False<br>提示：</p>
<p>字符串长度在[0, 100000]范围内。<br>说明:</p>
<p>你能只调用一次检查子串的方法吗？</p>
<figure class="highlight swift"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>{</span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">isFlipedString</span>(<span class="keyword">_</span> <span class="params">s1</span>: <span class="type">String</span>, <span class="keyword">_</span> <span class="params">s2</span>: <span class="type">String</span>)</span> -> <span class="type">Bool</span> {</span><br><span class="line"> <span class="keyword">if</span> s1 <span class="operator">==</span> <span class="string">""</span> <span class="operator">&&</span> s2 <span class="operator">==</span> <span class="string">""</span> { <span class="keyword">return</span> <span class="literal">true</span> } <span class="comment">//处理空字符串</span></span><br><span class="line"> <span class="keyword">guard</span> s1.count <span class="operator">==</span> s2.count <span class="keyword">else</span> {<span class="keyword">return</span> <span class="literal">false</span>}</span><br><span class="line"> <span class="keyword">return</span> (s1 <span class="operator">+</span> s1).contains(s2)</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>解题思路：</p>
<ol>
<li>利用s1+s1拼接起来，这样子就包括所有的旋转字符串了</li>
<li>只要新的字符串包含s2直接返回true</li>
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<h3 id="1-最小栈"><a href="#1-最小栈" class="headerlink" title="1.最小栈"></a>1.<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/min-stack/">最小栈</a></h3><p>设计一个支持 <code>push</code> ，<code>pop</code> ，<code>top</code> 操作，并能在常数时间内检索到最小元素的栈。</p>
<p>push(x) —— 将元素 x 推入栈中。<br>pop() —— 删除栈顶的元素。<br>top() —— 获取栈顶元素。<br>getMin() —— 检索栈中的最小元素。</p>
<p>示例:</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><span class="line">输入：</span><br><span class="line">["MinStack","push","push","push","getMin","pop","top","getMin"]</span><br><span class="line">[[],[-2],[0],[-3],[],[],[],[]]</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">输出：</span><br><span class="line">[null,null,null,null,-3,null,0,-2]</span><br><span class="line"></span><br><span class="line">解释：</span><br><span class="line">MinStack minStack = new MinStack();</span><br><span class="line">minStack.push(-2);</span><br><span class="line">minStack.push(0);</span><br><span class="line">minStack.push(-3);</span><br><span class="line">minStack.getMin(); --> 返回 -3.</span><br><span class="line">minStack.pop();</span><br><span class="line">minStack.top(); --> 返回 0.</span><br><span class="line">minStack.getMin(); --> 返回 -2.</span><br><span class="line"></span><br></pre></td></tr></table></figure>
<p>提示：</p>
<p><code>pop、top 和 getMin 操作总是在 非空栈 上调用。</code></p>
<figure class="highlight swift"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">MinStack</span> </span>{</span><br><span class="line"></span><br><span class="line"> <span class="keyword">var</span> head:<span class="type">ListNode</span>?</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">init</span>()</span> {</span><br><span class="line"> head <span class="operator">=</span> <span class="type">ListNode</span>(min:<span class="type">Int</span>.max)</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">push</span>(<span class="keyword">_</span> <span class="params">val</span>: <span class="type">Int</span>)</span> {</span><br><span class="line"> <span class="comment">//每次push都记录最小值</span></span><br><span class="line"> head <span class="operator">=</span> <span class="type">ListNode</span>(val: val, min: <span class="built_in">min</span>(val, head<span class="operator">!</span>.min), next: head)</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">pop</span>()</span> {</span><br><span class="line"> head <span class="operator">=</span> head<span class="operator">?</span>.next</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">top</span>()</span> -> <span class="type">Int</span> {</span><br><span class="line"> <span class="keyword">return</span> head<span class="operator">?</span>.val <span class="operator">??</span> <span class="number">0</span></span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> <span class="function"><span class="keyword">func</span> <span class="title">getMin</span>()</span> -> <span class="type">Int</span> {</span><br><span class="line"> <span class="keyword">return</span> head<span class="operator">?</span>.min <span class="operator">??</span> <span class="number">0</span></span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> <span class="class"><span class="keyword">class</span> <span class="title">ListNode</span> </span>{</span><br><span class="line"> <span class="keyword">let</span> val: <span class="type">Int</span></span><br><span class="line"> <span class="keyword">let</span> min:<span class="type">Int</span></span><br><span class="line"> <span class="keyword">let</span> next: <span class="type">ListNode</span>?</span><br><span class="line"></span><br><span class="line"> <span class="function"><span class="keyword">init</span>(<span class="params">val</span>: <span class="type">Int</span> <span class="operator">=</span> <span class="number">0</span>,<span class="params">min</span>:<span class="type">Int</span> <span class="operator">=</span> <span class="number">0</span>,<span class="params">next</span>:<span class="type">ListNode</span>? <span class="operator">=</span> <span class="literal">nil</span>)</span> {</span><br><span class="line"> <span class="keyword">self</span>.val <span class="operator">=</span> val</span><br><span class="line"> <span class="keyword">self</span>.min <span class="operator">=</span> min</span><br><span class="line"> <span class="keyword">self</span>.next <span class="operator">=</span> next</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"></span><br><span class="line"><span class="comment">/**</span></span><br><span class="line"><span class="comment"> * Your MinStack object will be instantiated and called as such:</span></span><br><span class="line"><span class="comment"> * let obj = MinStack()</span></span><br><span class="line"><span class="comment"> * obj.push(val)</span></span><br><span class="line"><span class="comment"> * obj.pop()</span></span><br><span class="line"><span class="comment"> * let ret_3: Int = obj.top()</span></span><br><span class="line"><span class="comment"> * let ret_4: Int = obj.getMin()</span></span><br><span class="line"><span class="comment"> */</span></span><br></pre></td></tr></table></figure>
<p>解题思路：</p>
<p>题目要求我们从栈元素中找到最小值，按照正常想，我们都会遍历一遍栈的元素，但是这样时间复杂度就是O(n),题目要求是O(1)，一般遇到这种问题，就需要空间换时间了。<br>使用链表，在每个节点保存当前的最小值，每次push都记录当前元素对应的最小值，每个节点都一一对应一个最小值，这样子getMin时间复杂度就是O(1)，同道理，这里用两个栈也可以做到一一对应也能解决。</p>
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<h3 id="1-移除链表元素"><a href="#1-移除链表元素" class="headerlink" title="1.移除链表元素"></a>1.<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/remove-linked-list-elements/">移除链表元素</a></h3><p>给你一个链表的头节点 <code>head</code> 和一个整数 <code>val</code> ，请你删除链表中所有满足 <code>Node.val == val</code> 的节点，并返回 <strong>新的头节点</strong> 。</p>
<p>**示例 **</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：head = [1,2,6,3,4,5,6], val = 6</span><br><span class="line">输出：[1,2,3,4,5]</span><br></pre></td></tr></table></figure>
<figure class="highlight swift"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">ListNode</span> </span>{</span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">var</span> val: <span class="type">Int</span></span><br><span class="line"> <span class="keyword">public</span> <span class="keyword">var</span> next: <span class="type">ListNode</span>?</span><br><span class="line"> <span class="keyword">public</span> <span class="function"><span class="keyword">init</span>()</span> { <span class="keyword">self</span>.val <span class="operator">=</span> <span class="number">0</span>; <span class="keyword">self</span>.next <span class="operator">=</span> <span class="literal">nil</span>; }</span><br><span class="line"> <span class="keyword">public</span> <span class="function"><span class="keyword">init</span>(<span class="keyword">_</span> <span class="params">val</span>: <span class="type">Int</span>)</span> { <span class="keyword">self</span>.val <span class="operator">=</span> val; <span class="keyword">self</span>.next <span class="operator">=</span> <span class="literal">nil</span>; }</span><br><span class="line"> <span class="keyword">public</span> <span class="function"><span class="keyword">init</span>(<span class="keyword">_</span> <span class="params">val</span>: <span class="type">Int</span>, <span class="keyword">_</span> <span class="params">next</span>: <span class="type">ListNode</span>?)</span> { <span class="keyword">self</span>.val <span class="operator">=</span> val; <span class="keyword">self</span>.next <span class="operator">=</span> next; }</span><br><span class="line">}</span><br><span class="line"><span class="function"><span class="keyword">func</span> <span class="title">removeElements</span>(<span class="keyword">_</span> <span class="params">head</span>: <span class="type">ListNode</span>?, <span class="keyword">_</span> <span class="params">val</span>: <span class="type">Int</span>)</span> -> <span class="type">ListNode</span>? {</span><br><span class="line"> <span class="keyword">guard</span> head <span class="operator">!=</span> <span class="literal">nil</span> <span class="keyword">else</span> { <span class="keyword">return</span> head }</span><br><span class="line"> <span class="comment">//虚拟头结点 也是新链表的头结点</span></span><br><span class="line"> <span class="keyword">let</span> dummyHead <span class="operator">=</span> <span class="type">ListNode</span>() </span><br><span class="line"> <span class="comment">//新链表尾结点</span></span><br><span class="line"> <span class="keyword">var</span> newTail <span class="operator">=</span> dummyHead</span><br><span class="line"> </span><br><span class="line"> <span class="comment">//用来遍历的链表头，swift传参都是不可变的</span></span><br><span class="line"> <span class="keyword">var</span> curHead <span class="operator">=</span> head</span><br><span class="line"></span><br><span class="line"> <span class="comment">//遍历链表</span></span><br><span class="line"> <span class="keyword">while</span> curHead <span class="operator">!=</span> <span class="literal">nil</span> {</span><br><span class="line"> <span class="keyword">if</span> curHead<span class="operator">?</span>.val <span class="operator">!=</span> val {</span><br><span class="line"> newTail.next <span class="operator">=</span> curHead</span><br><span class="line"> newTail <span class="operator">=</span> curHead<span class="operator">!</span></span><br><span class="line"> }</span><br><span class="line"> curHead <span class="operator">=</span> curHead<span class="operator">?</span>.next</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">//尾部置空</span></span><br><span class="line"> newTail.next <span class="operator">=</span> <span class="literal">nil</span></span><br><span class="line"></span><br><span class="line"> <span class="keyword">return</span> dummyHead.next</span><br><span class="line"> }</span><br></pre></td></tr></table></figure>
<p>解题思路：<br>这里用一个虚拟头结点来作为新链表的头结点会让代码更加的简洁，遍历传进来的链表，判断当前的链表是否是需要删除的如果不是需要删除的，让新链表尾结点下一个指向它，也就是尾部结点也是当前结点，最后把尾部结点置空，返回新节点头的下一个结点。</p>
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<h3 id="1-合并两个有序数组"><a href="#1-合并两个有序数组" class="headerlink" title="1. 合并两个有序数组"></a>1. <a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/merge-sorted-array/">合并两个有序数组</a></h3><p>给你两个按 非递减顺序 排列的整数数组 nums1 和 nums2，另有两个整数 m 和 n ，分别表示 nums1 和 nums2 中的元素数目。</p>
<p>请你 合并 nums2 到 nums1 中，使合并后的数组同样按 非递减顺序 排列。</p>
<p>注意：最终，合并后数组不应由函数返回，而是存储在数组 nums1 中。为了应对这种情况，nums1 的初始长度为 m + n，其中前 m 个元素表示应合并的元素，后 n 个元素为 0 ，应忽略。nums2 的长度为 n 。</p>
<p><strong>示例 ：</strong></p>
<p>输入：nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3<br>输出：[1,2,2,3,5,6]<br>解释：需要合并 [1,2,3] 和 [2,5,6] 。<br>合并结果是 [1,2,2,3,5,6] ，其中斜体加粗标注的为 nums1 中的元素。</p>
<figure class="highlight swift"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="keyword">func</span> <span class="title">merge</span>(<span class="keyword">_</span> <span class="params">nums1</span>: <span class="keyword">inout</span> [<span class="type">Int</span>], <span class="keyword">_</span> <span class="params">m</span>: <span class="type">Int</span>, <span class="keyword">_</span> <span class="params">nums2</span>: [<span class="type">Int</span>], <span class="keyword">_</span> <span class="params">n</span>: <span class="type">Int</span>)</span> {</span><br><span class="line"> <span class="comment">//思路是使用双指针的思路，这里用到了归并排序的思想</span></span><br><span class="line"> <span class="keyword">var</span> i1 <span class="operator">=</span> m <span class="operator">-</span> <span class="number">1</span></span><br><span class="line"> <span class="keyword">var</span> i2 <span class="operator">=</span> n <span class="operator">-</span> <span class="number">1</span></span><br><span class="line"> <span class="keyword">var</span> cur <span class="operator">=</span> nums1.count <span class="operator">-</span> <span class="number">1</span></span><br><span class="line"> <span class="keyword">while</span> i2 <span class="operator">>=</span> <span class="number">0</span> { <span class="comment">//这里必须要大于等于0 代表还没有合并结束</span></span><br><span class="line"> <span class="keyword">if</span> i1 <span class="operator">>=</span> <span class="number">0</span> <span class="operator">&&</span> nums2[i2] <span class="operator"><</span> nums1[i1] {</span><br><span class="line"> nums1[cur] <span class="operator">=</span> nums1[i1]</span><br><span class="line"> i1 <span class="operator">-=</span> <span class="number">1</span></span><br><span class="line"> }<span class="keyword">else</span>{ <span class="comment">// i1 < 0 || nums2[i2] >= nums1[i1] 这里i1已经为0了</span></span><br><span class="line"> nums1[cur] <span class="operator">=</span> nums2[i2]</span><br><span class="line"> i2 <span class="operator">-=</span> <span class="number">1</span></span><br><span class="line"> }</span><br><span class="line"> cur <span class="operator">-=</span> <span class="number">1</span></span><br><span class="line"> }</span><br><span class="line"> }</span><br></pre></td></tr></table></figure>
<p>解题思路：</p>
<p>双指针法，从后面插入替换合并，把nums2数值添加到nums1，如果在前面插入会替换到后面的数字</p>
<p>只要nums2的指针索引>=0代表还没有合并结束，只要成功合并一次，两个指针索引都需要向后移动一位</p>
<p>这里还需要考虑到比如nums1的长度比nums2 数组长度短的情况，所以 else 里就是 // i1 < 0 || nums2[i2] >= nums1[i1] 这里i1已经为0了 这里直接把nums2的数字往nums1数组合并</p>
<p>这里解题思路用到了归并排序算法：</p>
<p><img src="https://cdn.jsdelivr.net/gh/Tate-zwt/ImageHost/blog/202112241037111.gif" alt="图片"></p>
<ol>
<li>申请空间，使其大小为两个已经排序序列之和，该空间用来存放合并后的序列；</li>
<li>设定两个指针，最初位置分别为两个已经排序序列的起始位置；</li>
<li>比较两个指针所指向的元素，选择相对小的元素放入到合并空间，并移动指针到下一位置；</li>
<li>重复步骤 3 直到某一指针达到序列尾；</li>
<li>将另一序列剩下的所有元素直接复制到合并序列尾。</li>
</ol>
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<h2 id="初识"><a href="#初识" class="headerlink" title="初识"></a>初识</h2><p>作为一个移动端iOS开发，小程序也是有必要学习的，学好小程序的基本流程，会发现跟iOS原生开发很多都相似的，比如每个界面都有它的生命周期，数据绑定，响应事件等等，而且微信团队把它封装得非常易用，只要看文档，补充下CSS的知识就可以上手开发。</p>
<h3 id="CSS布局知识"><a href="#CSS布局知识" class="headerlink" title="CSS布局知识"></a>CSS布局知识</h3><p>clssname .clssname</p>
<figure class="highlight css"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"><span class="selector-class">.clssname</span> <span class="selector-class">.clssname</span>{</span><br><span class="line"> <span class="attribute">float</span><span class="selector-pseudo">:left</span>;</span><br><span class="line"> <span class="attribute">height</span>: <span class="number">100</span>rpx;</span><br><span class="line"> <span class="attribute">line-height</span>: <span class="number">100</span>rpx;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>代表 clssname 里的clssname样式更改</p>
<p>clssname view</p>
<figure class="highlight css"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="selector-class">.clssname</span> view{</span><br><span class="line"> <span class="attribute">flex</span>:<span class="number">1</span>;</span><br><span class="line"> <span class="attribute">color</span>:white;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>代表 clssname 里所有的view样式更改</p>
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<h3 id="二分查找算法-非递归-："><a href="#二分查找算法-非递归-：" class="headerlink" title="二分查找算法(非递归)："></a>二分查找算法(非递归)：</h3><ol>
<li>二分查找算法有递归和非递归方式，下面我们讲解二分查找算法的非递归方式 </li>
<li>二分查找法只适用于从有序的数列中进行查找(比如数字和字母等)，将数列排序后再进行查找</li>
<li>二分查找法的运行时间为对数时间 O(㏒₂n) ，即查找到需要的目标位置最多只需要㏒₂n 步，假设从[0,99]的 队列(100 个数，即 n=100)中寻到目标数 30，则需要查找步数为㏒₂100 , 即最多需要查找 7 次( 2^6 < 100 < 2^7)</li>
</ol>
<h3 id="分治算法："><a href="#分治算法：" class="headerlink" title="分治算法："></a>分治算法：</h3><ol>
<li>分治法是一种很重要的算法。字面上的解释是“分而治之”，就是把一个复杂的问题分成两个或更多的相同或 相似的子问题，再把子问题分成更小的子问题……直到最后子问题可以简单的直接求解，原问题的解即子问题 的解的合并。这个技巧是很多高效算法的基础，如排序算法(快速排序，归并排序)，傅立叶变换(快速傅立叶变 换)…… </li>
<li>分治算法可以求解的一些经典问题</li>
</ol>
<ul>
<li>二分搜索 </li>
<li>大整数乘法 </li>
<li>棋盘覆盖 </li>
<li>合并排序 </li>
<li>快速排序 </li>
<li>线性时间选择</li>
<li>最接近点对问题 </li>
<li>循环赛日程表 </li>
<li>汉诺塔</li>
</ul>
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<h2 id="1、使用SwiftRealm会导致没法预览报Updating-took-more-than-5-seconds"><a href="#1、使用SwiftRealm会导致没法预览报Updating-took-more-than-5-seconds" class="headerlink" title="1、使用SwiftRealm会导致没法预览报Updating took more than 5 seconds"></a>1、使用SwiftRealm会导致没法预览报<code>Updating took more than 5 seconds</code></h2><p>花了点时间查找问题，最后发现是因为初始化数据库写在AppDelegate，写在SceneDelegate就没事了！</p>
<h2 id="2、SwiftUI数据流温故知新-属性包装"><a href="#2、SwiftUI数据流温故知新-属性包装" class="headerlink" title="2、SwiftUI数据流温故知新 - 属性包装"></a>2、SwiftUI数据流温故知新 - 属性包装</h2><h3 id="State"><a href="#State" class="headerlink" title="@State"></a>@State</h3><p>@State属性允许修改Struct的属性，这些属性在普通的Struct里面是不允许修改的。当把@State放置到属性前，该属性实际上会被放到Struct的外部存储起来，这意味着SwiftUI能够随时销毁和重建Struct而不会丢失属性的值。<br>@State包装的属性通常是设置成私有的，不让外部使用。如果想让外部使用，则应该使用@ObservedObject和@EnvironmentObject，他们能够使外部修改属性后，状态能够得到改变。<br>建议把@State包装的属性都设置成私有</p>
<h3 id="Published"><a href="#Published" class="headerlink" title="@Published"></a>@Published</h3><p>@Published是SwiftUI最有用的包装之一，允许我们创建出能够被自动观察的对象属性，SwiftUI会自动监视这个属性，一旦发生了改变，会自动修改与该属性绑定的界面。<br>1.首先需要遵循ObservableObject属性</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">class Bag: ObservableObject { var items = [String]() }</span><br></pre></td></tr></table></figure>
<p>2.包装属性</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">class Bag: ObservableObject { @Published var items = [String]() }</span><br></pre></td></tr></table></figure>
<p>这样就完成了。@Published包装会自动添加willSet方法监视属性的改变。</p>
<h3 id="ObservedObject"><a href="#ObservedObject" class="headerlink" title="@ObservedObject"></a>@ObservedObject</h3><p>@ObservedObject告诉SwiftUI，这个对象是可以被观察的，里面含有被@Published包装了的属性。<br>@ObservedObject包装的对象，必须遵循ObservableObject协议。也就是说必须是class对象，不能是struct。<br>@ObservedObject允许外部进行访问和修改。</p>
<h3 id="EnvironmentObject"><a href="#EnvironmentObject" class="headerlink" title="@EnvironmentObject"></a>@EnvironmentObject</h3><p>@EnvironmentObject包装的属性是全局的，整个app都可以访问<br>需要注意的是，不需要给定默认值，由于该属性是整个app都可以访问的，SwiftUI会自动的从环境中取出来。<br>当有这样一个场景，A->B->C->D->E->F，A界面的数据要传递给F界面，假如使用@ObservedObject包装，则需要一层一层传递，而使用@EnvironmentObject则不需要，直接在F界面，通过SwiftUI环境直接取出来就行。<br>需要注意的是，当界面显示时，就会去环境中取，但是，假如之前没有把属性放到环境中，则程序会崩溃。</p>
<h3 id="Environment"><a href="#Environment" class="headerlink" title="@Environment"></a>@Environment</h3><p>@Environment与@EnvironmentObject作用是不同的，@Environment是从环境中取出预定义的值，比如获得当前是暗黑模式还是正常模式，屏幕的大小等等。</p>
<figure class="highlight plaintext"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">@Environment(\.horizontalSizeClass) var horizontalSizeClass</span><br><span class="line">@Environment(\.managedObjectContext) var managedObjectContext</span><br></pre></td></tr></table></figure>
<h3 id="Binding"><a href="#Binding" class="headerlink" title="@Binding"></a>@Binding</h3><p>@Binding也是非常重要的一个包装，声明一个属性是从外部获取的，并且与外部是共享的。相当于外部传过来的时候，不是传递的值。任何一方修改都会让SwiftUI的监视生效。</p>
<h3 id="GestureState"><a href="#GestureState" class="headerlink" title="@GestureState"></a>@GestureState</h3><p>@GestureState能够让我们传递手势的状态，虽然使用@State也能实现，但@GestureState能让手势结束后我们回到最初的状态。</p>
<p>设置属性，设置zero意味着手势结束后，会回到最初的值。</p>
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