From 46a6f2851c79de41d4089949e58a6376774cd37b Mon Sep 17 00:00:00 2001
From: humingk <humingk@qq.com>
Date: Thu, 13 Jun 2024 22:36:16 +0800
Subject: [PATCH] add

---
 _posts/2019-03-18-alg-note.md | 115 +++++++++++++++++++++++++++++++++-
 1 file changed, 112 insertions(+), 3 deletions(-)

diff --git a/_posts/2019-03-18-alg-note.md b/_posts/2019-03-18-alg-note.md
index 3683fe6..4432f2d 100644
--- a/_posts/2019-03-18-alg-note.md
+++ b/_posts/2019-03-18-alg-note.md
@@ -891,8 +891,10 @@ class Solution {
 
 ![](https://raw.githubusercontent.com/humingk/resource/master/image/2019/sort_quick_partition1.png)
 
-前后遍历双切分：
+###### 前后遍历双切分（推荐）：
+
 https://leetcode.cn/problems/sort-an-array/
+
 ```java
     public int[] sortArray(int[] nums) {
         if (nums == null || nums.length <= 1) {
@@ -936,7 +938,7 @@ https://leetcode.cn/problems/sort-an-array/
     }
 ```
 
-**前后遍历三切分（推荐）：**
+###### **前后遍历三切分（推荐）：**
 
 ```java
     public int[] sortArray(int[] nums) {
@@ -6007,7 +6009,7 @@ PS：
 
 ---
 
-#### 解法2 类快速排序 O(N) + O(1)
+#### 解法2 类快速排序 O(NLogN) + O(1)
 
 比第k个数字小的所有数字都在k的左边，比第k个数字大的都在k的右边
 
@@ -6074,6 +6076,61 @@ public class Solution {
     }
 }
 ```
+oj：https://leetcode.cn/problems/smallest-k-lcci/
+另一种写法
+
+```java
+class Solution {
+    public int[] smallestK(int[] arr, int k) {
+        quickSort(arr, 0, arr.length - 1, k);
+        int[] result = new int[k];
+        for (int i = 0; i < k; i++) {
+            result[i] = arr[i];
+        }
+        return result;
+    }
+
+    private void quickSort(int[] arr, int min, int max, int k) {
+        if (min >= max) {
+            return;
+        }
+        // 拆分
+        int target = arr[min];
+        int start = min, end = max + 1;
+        while (true) {
+            while (start < max && arr[++start] < target)
+                ;
+            while (end > min && arr[--end] > target)
+                ;
+            if (start < end) {
+                exchange(arr, start, end);
+            } else {
+                break;
+            }
+        }
+        exchange(arr, min, end);
+        // end左边的都小，end右边的都大，直接返回左边即可
+        if (end == k - 1) {
+            return;
+        }
+        // end左边的有小有大，继续拆分左边
+        else if (end > k - 1) {
+            quickSort(arr, min, end - 1, k);
+        }
+        // end右边的有大有小，继续拆分右边
+        else {
+            quickSort(arr, end + 1, max, k);
+        }
+    }
+
+    private void exchange(int[] arr, int left, int right) {
+        int temp = arr[left];
+        arr[left] = arr[right];
+        arr[right] = temp;
+    }
+}
+```
+
 
 ------
 
@@ -6184,6 +6241,9 @@ class Solution {
 
 
 
+
+
+
 ---
 
 ### 连续子数组的最大和
@@ -6350,6 +6410,55 @@ public class Solution {
 }
 ```
 
+oj:https://leetcode.cn/problems/shu-ju-liu-zhong-de-zhong-wei-shu-lcof/
+也可以通过判断两个队列大小是否相等来平分
+
+```java
+class MedianFinder {
+
+    private PriorityQueue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> o2 - o1);
+    private PriorityQueue<Integer> minHeap = new PriorityQueue<>();
+
+    /** initialize your data structure here. */
+    public MedianFinder() {
+
+    }
+
+    public void addNum(int num) {
+        // 平分插入堆
+        if(minHeap.size()==maxHeap.size()){
+            maxHeap.add(num);
+            // 最大堆的最大值插入最小堆
+            minHeap.add(maxHeap.poll());
+        }else{
+            minHeap.add(num);
+            // 最小堆的最小值插入最大堆
+            maxHeap.add(minHeap.poll());
+        }
+    }
+
+    public double findMedian() {
+        // 最小堆和最大堆个数一样，返回平均值
+        if(minHeap.size()==maxHeap.size()){
+            return (minHeap.peek()+maxHeap.peek())/(double)2; 
+        }
+        // 最小堆和最大堆个数不一样，返回中位数
+        else if(minHeap.size()>maxHeap.size()){
+            return minHeap.peek();
+        }else{
+            return maxHeap.peek();
+        }
+    }
+}
+
+/**
+ * Your MedianFinder object will be instantiated and called as such:
+ * MedianFinder obj = new MedianFinder();
+ * obj.addNum(num);
+ * double param_2 = obj.findMedian();
+ */
+```
+
 ------
 
 ### 整数中1出现的次数