From 8b107b4cf9368c203a492eccc4280e6dc8934771 Mon Sep 17 00:00:00 2001
From: lrc123 <892834414@github.com>
Date: Fri, 1 May 2020 20:07:57 +0800
Subject: [PATCH 1/4] lrc123 first commit
---
Heap/01-introduction.md | 44 +++++++++
Heap/02-template-1.md | 178 +++++++++++++++++++++++++++++++++++++
Heap/03-template-2.md | 39 ++++++++
Heap/04-template-3.md | 148 ++++++++++++++++++++++++++++++
Heap/05-examples.md | 51 +++++++++++
Heap/06-practices.md | 20 +++++
Heap/images/add.png | Bin 0 -> 81619 bytes
Heap/images/array.png | Bin 0 -> 62566 bytes
Heap/images/binaryheap.png | Bin 0 -> 73592 bytes
_sidebar.md | 8 ++
10 files changed, 488 insertions(+)
create mode 100644 Heap/01-introduction.md
create mode 100644 Heap/02-template-1.md
create mode 100644 Heap/03-template-2.md
create mode 100644 Heap/04-template-3.md
create mode 100644 Heap/05-examples.md
create mode 100644 Heap/06-practices.md
create mode 100644 Heap/images/add.png
create mode 100644 Heap/images/array.png
create mode 100644 Heap/images/binaryheap.png
diff --git a/Heap/01-introduction.md b/Heap/01-introduction.md
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+# 堆数据结构和相关算法应用
+
+作者： Lrc123 审核:
+
+## 什么是堆?
+
+堆的性质其实很简单，将堆看作一棵树，那么该树总满足父节点的值大于其子节点(最大堆)；同理，最小堆是其父节点的值总小于其子节点。
+
+如此一来，从最大堆每次取出的堆顶元素是该堆中的最大元素，从最小堆每次取出的堆顶元素是该堆中的最小元素。
+
+## 二叉堆
+>「二叉堆（Binary Heap）是一个可以被看成近似完全二叉树的数组。树上的每一个结点对应数组的一个元素。除了最底层外，该树是完全充满的，而且是从左到右填充。」 —— 《算法导论》
+
+上面这段引用什么意思呢，用一句话概括就是：`二叉堆在逻辑上是一颗完全二叉树，在实现上是普通的一维数组。`
+
+我们用两幅图直观地感受一下:
+
+
+图1. 逻辑上的二叉树
+
+
+图2. 实现上的一维数组
+
+## 堆常见的操作：
+
+HEAPIFY 建堆：把一个无序的数组变成堆结构的数组，时间复杂度为$O(N\log N)$。
+
+HEAPPUSH：向数组末尾加入新的元素，并调整到正确的位置，维持堆的结构，时间复杂度为 $O(\log N)$。
+
+HEAPPOP：弹出堆顶元素，并调整堆，替代到堆顶位置，时间复杂度为 $O(\log N)$。
+
+HEAPSORT：HEAPFY 维持堆的结构，通过HEAPPOP，每次将最堆顶元素排列到堆尾，size减小。重复操作直到数组绝对有序，时间复杂度为$O(N\log N)$。空间复杂度为 $O(1)$。
+
+
+## 学习建议
++ 我们可以先通过手动实现一个堆的数据结构来认识堆。
++ 当我们遇到题目中有优先级、大小关系或排序等字眼的时候，我们就可以想到堆这个数据结构，并尝试应用它。
+
+# 堆的应用和刷题常见问题
++ 堆与优先队列
+ 堆与优先队列这个标题常常一起出现，实际上堆就是优先队列的一种实现方式。
+ 我们在实例化优先队列的时候，也常常将实例名称写为maxHeap和minHeap。
++ `注意：这里的堆指的是一种数据结构，不是操作系统里的堆。`
+
diff --git a/Heap/02-template-1.md b/Heap/02-template-1.md
new file mode 100644
index 0000000..58c4f73
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+++ b/Heap/02-template-1.md
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+# 堆的构建和堆排序
+
+在我们做算法题的大多数情况，不需要手动地构建堆这个数据结构。通常我们直接使用编程语言自带的实现，如java使用PriorityQueue。不过，在学习堆这个数据结构的时候，手动构建堆，可以加深理解，帮助我们更好地使用它。
+
+### 堆(最大堆)的构建
+
+1. 在介绍章节中我们讲过堆的构建要通过一维数组来实现，那么如何维持一维数组中的逻辑关系呢？这里用到3个辅助方法, 来计算父节点和子节点的位置关系。
+
+
+#### **Java**
+```java
+
+// 通过当前位置求父节点位置
+public int parent(int index){
+ return arr[(index - 1) / 2];
+}
+
+// 通过当前位置求左孩子节点位置
+public int leftChild(int index){
+ return arr[index * 2 + 1];
+}
+
+// 通过当前位置求右孩子节点位置
+public int rightChild(int index){
+ return arr[index * 2 + 2];
+}
+
+```
+
+
+2. 向堆中添加元素和SiftUp
+
+`SiftUp`操作是为了向堆中添加元素，和直接对数组`heapify`(将一个数组堆化)不同的是，`SiftUp`用于动态地构造一个堆。将从数组尾部添加的新元素给移动到正确的位置。
+
+
+#### **Java**
+```java
+
+/**
+* 先从数组尾部添加新元素，再将该元素移动到正确的位置。而这个移动的过程就叫做siftUp。
+*/
+public void add(int num){
+ data.add(num);
+ siftUp(num);
+}
+
+// siftUp 通过比较当前节点和父节点的大小，将较大的孩子节点交换到父亲节点的位置。同时不要忘记交换后节点位置也要更新为原父节点的位置。
+public void siftUp(int index){
+ while(data.get(index) > data.get(parent(index))){
+ swap(data, index, parent(index));
+ index = parent(index);
+ }
+}
+
+```
+
+
+3. 从堆中取出元素和SiftDown
+
+
+#### **Java**
+
+```java
+/**
+取出数组顶部元素，然后将顶部元素和数组最后一个元素swap，移除数组末尾的元素
+*/
+public int extractMax(){
+ int maxVal = data.get(0);
+ swap(data, 0, data.size() - 1);
+ data.remove(data.size() - 1);
+ return maxVal;
+}
+/**
+ *
+ * @param index
+ */
+public void siftDown(int index){
+ while(leftChild(index) < data.size()) {
+ int j = leftChild(index);
+ if (j + 1 < data.size() && data.get(j + 1) > data.get(j)) {
+ j++;
+ }
+ if (data.get(index) > data.get(j)) {
+ return;
+ }
+ swap(data, index, j);
+ index = j;
+ }
+}
+
+```
+
+
+
+
+## 堆排序
+
++ 堆排序的原理是：假如要对一个数组进行从小到大的排列，则对数组`heapify`成一个最大堆。然后，每次将堆顶元素放到数组末尾，使`size`减一。当每次取出的当前最大元素按照`size`递减的位置排列后，就得到了一个有序的数组。
+
+
+#### **Java**
+
+```java
+public class Solution {
+
+ /**
+ * 这里的heapsort是一种in-place排序，就是直接对原数组进行位置上的交换来排序
+ * @param nums
+ */
+ public void heapSort(int[] nums) {
+ int size = nums.length - 1;
+ // 先对数组进行最大堆化
+ heapify(nums, size);
+ // 然后每次将堆顶的最大元素移动到数组末尾，将交换至堆顶的较小元素再进行下沉操作，使数组再次堆化有序。使size减一，堆化的操作将不影响末尾已经排序好的元素。重复操作直到堆中元素都按照从大到小，从后往前的方式排列在原数组上。
+ while(size > 0){
+ swap(nums, 0, size);
+ siftDown(nums, 0, size);
+ size --;
+ }
+ }
+
+ //堆化，从末尾开始将较小元素下沉
+ public void heapify(int[] nums, int size){
+ for(int i = parent(size); i > 0; i--){
+ siftDown(nums, i, size);
+ }
+ }
+
+ //下沉操作
+ public void siftDown(int[] nums, int index, int size){
+ while(leftChild(index) < size){
+ // 这里声明的j变量是用来选择当前节点的两个左右孩子的较大节点
+ int j = leftChild(index);
+ if(j + 1 < size && nums[j + 1] > nums[j]){
+ j++;
+ }
+ // 如果两个孩子节点的较大值都比当前节点小，则不需要继续swap，直接return
+ if(nums[index] > nums[j]){
+ return;
+ }
+ // 如果较大的那个比当前节点大，则交换
+ swap(nums, index, j);
+ // 当前节点的index也要移动到被交换的孩子节点上，为下一次比较做准备
+ index = j;
+ }
+ }
+
+ // 对原数组进行操作，交换两个位置的变量
+ public void swap(int[] nums, int i, int j){
+ int tmp = nums[i];
+ nums[i] = nums[j];
+ nums[j] = tmp;
+ }
+
+ // 输入当前节点的位置，计算得到父节点的位置
+ public int parent(int index){
+ return (index - 1) / 2;
+ }
+
+ // 计算左孩子节点的位置
+ public int leftChild(int index){
+ return index * 2 + 1;
+ }
+
+ // 计算右孩子节点的位置
+ public int rightChild(int index){
+ return index * 2 + 2;
+ }
+}
+
+```
+
+
+
+**复杂度分析**：
+
++ 时间复杂度：$O(N\log N)$，这里 $N$ 是数组的长度，每次抽取出堆顶为$O(\log N)$，$N$个元素就N次操作。
++ 空间复杂度：$O(1)$。in-place排序，不需要额外空间。
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diff --git a/Heap/03-template-2.md b/Heap/03-template-2.md
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+# 求(前/第）K(大/小/个)问题
+
+当我们遇到题目中有优先级、大小关系或排序等字眼的时候，我们就可以想到堆这个数据结构。
+
+
+1. 第K大/小问题
+问题分析：这类问题求解的关键在于如何从无序的数组中，得到必须在有序前提下才能得到的第k个元素。
+使无序变为有序的最简单方法就是排序，但是要得到我们需要的解却并不需要绝对有序。而堆，就可以使数组呈现一种整体有序的效果，从而更高效地求解问题。
+
+例题 1: 「力扣」第 215 题」：[数组中的第K个最大元素(easy)](https://leetcode-cn.com/problems/kth-largest-element-in-an-array/)
+
++ 当我们求数组中第k大的数字的时候，首先会想到的是构建一个最大堆存储所有的元素，然后将前面（k-1）个元素给弹出。最后返回堆顶。这确实是一个可行的办法。不过，反直觉的是，我们可以构建一个最小堆来更快地得到第k大的数字。
+
++ 因为，第k大的元素同时也是前k个最大元素中的最小值。这对应了我们一开始说的，最大最小想到堆。我们可以构建一个容量为k的最小堆来存储数组中最大的k个元素：当元素的数量超过k时，我们将堆顶最小的元素弹出，弹出不需要的(arr.length - k)个元素，剩下的就是我们需要的最大的k个元素，而堆顶也正式整个数组的第k大元素。
+
+
+#### **Java**
+
+```java
+
+public int findKthLargest(int[] nums, int k) {
+ PriorityQueue minHeap = new PriorityQueue<>();
+
+ for (Integer num : nums) {
+ minHeap.add(num);
+ if(minHeap.size() > k){
+ minHeap.poll();
+ }
+ }
+ return minHeap.peek();
+}
+
+```
+
+
+**复杂度分析**：
+
++ 时间复杂度：$O(N\log N)$。
++ 空间复杂度：$O(k)$。
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diff --git a/Heap/04-template-3.md b/Heap/04-template-3.md
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+# 双堆法（大小堆法）
+
+双堆法用于解决的题型有：中位数问题。
+
+1. 中位数问题
+例题 1: 「力扣」第 295 题 」：[数据流的中位数(hard)](https://leetcode-cn.com/problems/find-median-from-data-stream/)
+
+`问题分析`：
+当需要求一组数的中位数的时候，我们需要先将该组数排序，然后找到中间位置。假如数组的长度是奇数，则返回中间那个数。假如数组长度为偶数，则返回中间两个数字和的一半。
+
+这是非常自然想到的思路。
+
+那么，如果这个数组是不断增加的呢？
+我们会想到，新元素插入到排序数组里，然后再计算中位数对应下标的位置，就可以得到结果了。
+这是没有错的。不过，我们同时知道，在一个有序数组中插入一个值，时间复杂度为$O(N)$。这是因为，数组中插入一个元素，要挪动插入位置之后的所有数字。我们插入N次，大概可以得到$O(N^2)$的复杂度。有没有每次插入更快一点的数据结构呢。很显然，就是堆。每次插入的时间复杂度为$O(logN)$，N次插入大约为$O(NlogN)$。
+
+双堆法的思想在于，将数组划分为大小相差不超过1的两个堆，保持这样的平衡。两个堆的顶部就会恰好是数组中间的值。
+
+### 求中位数
+基于双堆，我们可以很快写出求中位数的方法。
+
++ 首先，如果两个堆都为空，则返回0.0。然后，如果最小堆为空，（第二个if已经排除了均为空的情况），则返回最大堆。最后，（都不为空）如果size相等，就是偶数，不相等，就是奇数。返回对应的答案即可。
++ 注意，为偶数的情况，我们的中位数计算还是要使用经典的写法：先减后加来防止溢出。
+
+
+#### **Java**
+
+```java
+public double findMedian() {
+ if(maxHeap.isEmpty() && minHeap.isEmpty()){
+ return 0.0;
+ }
+ if(minHeap.isEmpty()){
+ return maxHeap.peek();
+ }
+ return maxHeap.size() == minHeap.size() ? maxHeap.peek() + (minHeap.peek() - maxHeap.peek()) / 2.0 : (double)maxHeap.peek();
+}
+```
+
+
+### 添加元素
+
++ 基于刚写好的findMedian()方法，我们的添加操作非常轻松。如果要添加的值大于原数组的中位数，我们就把它放到较大的那一组，也就是最小堆里面。如果小于或者等于中位数，就把它放到最大堆里面。
++ 这个时候，我们不要忘记了一件事。就是，每次添加了新的元素，我们的中位数在数组中的位置是会向右偏移的，所以我们要平衡一下两个堆里面元素的个数。
+
+
+#### **Java**
+
+```java
+public void addNum(int num) {
+ if(num <= findMedian()){
+ maxHeap.add(num);
+ }else{
+ minHeap.add(num);
+ }
+ balance();
+}
+```
+
+
++ 这个balance的过程非常简单，就是把多了元素的堆，堆顶元素给弹出加入到另一个堆中。即可。
++ 注意，我们这里的实现，默认是最大堆可以比最小堆多一个元素的。
+
+
+#### **Java**
+
+```java
+public void balance(){
+ if(maxHeap.isEmpty() && minHeap.isEmpty()){
+ return ;
+ }
+ while(minHeap.size() > maxHeap.size()){
+ maxHeap.add(minHeap.poll());
+ }
+ while(maxHeap.size() - minHeap.size() > 1){
+ minHeap.add(maxHeap.poll());
+ }
+}
+
+```
+
+
+### 完整代码
+
+#### **Java**
+
+```java
+class MedianFinder {
+
+ PriorityQueue minHeap = null;
+ PriorityQueue maxHeap = null;
+
+ /** initialize your data structure here. */
+ public MedianFinder() {
+ // java 中默认的实现是最小堆
+ minHeap = new PriorityQueue<>();
+ maxHeap = new PriorityQueue<>(new Comparator() {
+ @Override
+ public int compare(Integer i1, Integer i2) {
+ return i2.compareTo(i1);
+ }
+ });
+ }
+
+ public void addNum(int num) {
+ if(num <= findMedian()){
+ maxHeap.add(num);
+ }else{
+ minHeap.add(num);
+ }
+ balance();
+ }
+
+ public void balance(){
+ if(maxHeap.isEmpty() && minHeap.isEmpty()){
+ return ;
+ }
+ while(minHeap.size() > maxHeap.size()){
+ maxHeap.add(minHeap.poll());
+ }
+ while(maxHeap.size() - minHeap.size() > 1){
+ minHeap.add(maxHeap.poll());
+ }
+ }
+
+ public double findMedian() {
+ if(maxHeap.isEmpty() && minHeap.isEmpty()){
+ return 0.0;
+ }
+ if(minHeap.isEmpty()){
+ return maxHeap.peek();
+ }
+ return maxHeap.size() == minHeap.size() ? maxHeap.peek() + (minHeap.peek() - maxHeap.peek()) / 2.0 : (double)maxHeap.peek();
+ }
+
+ public static void main(String[] args) {
+ MedianFinder finder = new MedianFinder();
+ System.out.println(finder.findMedian());
+ finder.addNum(1);
+ finder.addNum(2);
+ finder.addNum(3);
+ finder.addNum(4);
+ System.out.println(finder.findMedian());
+ }
+}
+
+```
+
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diff --git a/Heap/05-examples.md b/Heap/05-examples.md
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+++ b/Heap/05-examples.md
@@ -0,0 +1,51 @@
+# 精选例题
+
+## 题型1： 前k个XX问题
+前k个元素的输出其实很简单，前k大的就构建最大堆输出k次。前k小的就构建最小堆输出k次。不过，在leetcode中，前k个XX问题常常和频率结合在一起，组成了一个先统计再排序的问题。
+
+
+例题 1: 「力扣」：[692.前 K 个高频单词 (medium)](https://leetcode-cn.com/problems/top-k-frequent-words/)
+
+思路： 先使用hashmap统计单词出现频率，再存入堆，每次输出堆顶。
+
+
+#### **Java**
+
+```java
+
+class Solution {
+ public List topKFrequent(String[] words, int k) {
+
+ HashMap map = new HashMap<>();
+
+ for (String word : words) {
+ map.put(word, map.getOrDefault(word, 0) + 1);
+ }
+
+ PriorityQueue heap = new PriorityQueue<>(
+ (a, b) -> map.get(b).equals(map.get(a)) ? b.compareTo(a) : map.get(a) - map.get(b)
+ );
+
+ map.keySet().forEach(item -> {
+ heap.add(item);
+ if (heap.size() > k)
+ heap.poll();
+ });
+
+ LinkedList res = new LinkedList<>();
+ LinkedList freq = new LinkedList<>();
+ while(!heap.isEmpty()){
+ freq.add(map.get(heap.peek()));
+ res.add(heap.poll());
+ }
+ for (Integer item : freq) {
+ System.out.println(item);
+ }
+ Collections.reverse(res);
+ return res;
+ }
+}
+```
+
+
+
diff --git a/Heap/06-practices.md b/Heap/06-practices.md
new file mode 100644
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+++ b/Heap/06-practices.md
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+# 精选练习
+
+## 题型一： 前k个类型的题目
+| 题目 | 提示 |
+| ------------------------------------------------------------ | -------------------------------------- |
+| [215. 数组中的第K个最大元素 (easy)](https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/)（必做） | 例题 |
+| [264. 丑数 II (medium)](https://leetcode-cn.com/problems/ugly-number-ii/) | |
+| [347. 前 K 个高频元素 (medium)](https://leetcode-cn.com/problems/top-k-frequent-elements/) (必做) | 这类题型的基础题 |
+| [692. 前 K 个高频单词 (medium)](https://leetcode-cn.com/problems/top-k-frequent-words/) | |
+| [407. 接雨水 II (hard)](https://leetcode-cn.com/problems/trapping-rain-water-ii/) | |
+| [502. IPO(hard)](https://leetcode-cn.com/problems/ipo/) | 看起来难，用堆做简单 |
+
+
+## 题型二： 中位数类型的题目
+
+| 题目 | 提示 |
+| ------------------------------------------------------------ | ------------------ |
+| [4. 寻找两个有序数组的中位数(hard)](https://leetcode-cn.com/problems/median-of-two-sorted-arrays/) | 挺难的，官方题解用递归双指针啥的比较复杂，不过这里亲测直接用双堆法也可以过 |
+| [295. 数据流的中位数(hard)](https://leetcode-cn.com/problems/find-median-from-data-stream/) (必做)| 例题必做 |
+| [480. 滑动窗口的中位数(hard)](https://leetcode-cn.com/problems/sliding-window-median/) | 这题是数据流那题的进阶版，很好的练手题 |
\ No newline at end of file
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