diff --git a/Leecode/src/main/java/com/markilue/leecode/greedy/T10_1_ReconstructQueue.java b/Leecode/src/main/java/com/markilue/leecode/greedy/T10_1_ReconstructQueue.java
new file mode 100644
index 0000000..8cad6cb
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/greedy/T10_1_ReconstructQueue.java
@@ -0,0 +1,148 @@
+package com.markilue.leecode.greedy;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.LinkedList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.greedy
+ *@Author: markilue
+ *@CreateTime: 2023-02-10  11:01
+ *@Description:
+ * TODO  力扣406题  根据身高重建队列:
+ *  假设有打乱顺序的一群人站成一个队列，数组 people 表示队列中一些人的属性（不一定按顺序）。
+ *  每个 people[i] = [hi, ki] 表示第 i 个人的身高为 hi ，前面 正好 有 ki 个身高大于或等于 hi 的人。
+ *  请你重新构造并返回输入数组 people 所表示的队列。
+ *  返回的队列应该格式化为数组 queue ，其中 queue[j] = [hj, kj] 是队列中第 j 个人的属性（queue[0] 是排在队列前面的人）。
+ *@Version: 1.0
+ *
+ */
+public class T10_1_ReconstructQueue {
+
+    @Test
+    public void test() {
+//        int[][] people = new int[][]{{7, 0}, {4, 4}, {7, 1}, {5, 0}, {6, 1}, {5, 2},{4,1}};
+        int[][] people = new int[][]{{6, 0}, {5, 0}, {4, 0}, {3, 2}, {2, 2}, {1, 4}};
+        reconstructQueue(people);
+    }
+
+
+    /**
+     * 思路:先升序排列一下,再反向腾挪位置
+     * 速度击败78.37%，内存击败73.68%  6ms
+     * @param people
+     * @return
+     */
+    public int[][] reconstructQueue(int[][] people) {
+
+        //升序排列一下
+        Arrays.sort(people, (array1, array2) -> {
+            return array1[0] != array2[0] ? array1[0] - array2[0] : array1[1] - array2[1];
+        });
+
+        //反向遍历，挪位置
+        for (int i = people.length - 1; i >= 0; i--) {
+            //统计前面有多少个相等的数，看看这个值与people[i][1]是否相等
+            int count = 0;
+            for (int j = i - 1; j >= 0; j--) {
+                if (people[j][0] == people[i][0]) {
+                    count++;
+                } else break;
+            }
+            //不等于就需要向后挪动
+            if (people[i][1] != count) {
+                int[] cur = people[i];
+                int index = people[i][1] - count;
+                if (i + index < people.length) {
+                    System.arraycopy(people, i + 1, people, i, index);
+                    people[i + index] = cur;
+                }
+            }
+        }
+        return people;
+    }
+
+
+    /**
+     * 代码随想录思路:本质上一致，就是分别考虑两个维度，想考虑身高，在考虑位置
+     * 不过与本人不同的是：他是先由大到小排列，然后使用linkedList一个一个插入
+     * 时间击败67.5%   内存击败47.51%  7ms
+     * @param people
+     * @return
+     */
+    public int[][] reconstructQueue1(int[][] people) {
+        // 身高从大到小排（身高相同k小的站前面）
+        Arrays.sort(people, (a, b) -> {
+            if (a[0] == b[0]) return a[1] - b[1];
+            return b[0] - a[0];
+        });
+
+        LinkedList<int[]> que = new LinkedList<>();
+
+        for (int[] p : people) {
+            que.add(p[1],p);
+        }
+
+        return que.toArray(new int[people.length][]);
+    }
+
+
+    /**
+     * 官方最快:2ms
+     * 本质上只是自己实现了快排来进行排序
+     * @param people
+     * @return
+     */
+    public int[][] reconstructQueue2(int[][] people) {
+        int len = people.length;
+        quickSort(people, 0, len - 1);
+        List<int[]> ans = new ArrayList<int[]>();
+        for (int[] person : people) {
+            ans.add(person[1], person);
+        }
+        return ans.toArray(new int[ans.size()][]);
+    }
+
+    private void quickSort(int[][] array, int left, int right){
+        if(left >= right){
+            return;
+        }
+        int i = left, j = right;
+        int[] temp = array[left];
+        int index = left;
+        while(i < j){
+            while(i < j && compareValue1(array[i], temp)){
+                i++;
+            }
+            while(i < j && !compareValue1(array[j], temp)){
+                j--;
+            }
+            if(i < j){
+                int[] t = array[i];
+                array[i] = array[j];
+                array[j] = t;
+            }
+            if(array[i] == temp){
+                index = i;
+            }else if(array[j] == temp){
+                index = j;
+            }
+        }
+        array[index] = array[i];
+        array[i] = temp;
+        quickSort(array, left, i - 1);
+        quickSort(array, i + 1, right);
+    }
+
+    private boolean compareValue1(int[] a, int[] b){
+        if(a[0] == b[0]){
+            return a[1] < b[1];
+        }else{
+            return a[0] > b[0];
+        }
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/greedy/LemonadeChange.java b/Leecode/src/main/java/com/markilue/leecode/greedy/T10_LemonadeChange.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/greedy/LemonadeChange.java
rename to Leecode/src/main/java/com/markilue/leecode/greedy/T10_LemonadeChange.java
index cf617e0..839a7b1 100644
--- a/Leecode/src/main/java/com/markilue/leecode/greedy/LemonadeChange.java
+++ b/Leecode/src/main/java/com/markilue/leecode/greedy/T10_LemonadeChange.java
@@ -17,7 +17,7 @@ import java.util.HashMap;
  *  给你一个整数数组 bills ，其中 bills[i] 是第 i 位顾客付的账。如果你能给每位顾客正确找零，返回 true ，否则返回 false 。
  * @Version: 1.0
  */
-public class LemonadeChange {
+public class T10_LemonadeChange {
 
     @Test
     public void test(){
diff --git a/Leecode/src/main/java/com/markilue/leecode/greedy/second/T09_Candy.java b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T09_Candy.java
index baa1075..c041186 100644
--- a/Leecode/src/main/java/com/markilue/leecode/greedy/second/T09_Candy.java
+++ b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T09_Candy.java
@@ -1,5 +1,9 @@
 package com.markilue.leecode.greedy.second;
 
+import org.junit.Test;
+
+import java.util.Arrays;
+
 /**
  *@BelongsProject: Leecode
  *@BelongsPackage: com.markilue.leecode.greedy.second
@@ -16,7 +20,47 @@ package com.markilue.leecode.greedy.second;
  */
 public class T09_Candy {
 
+    @Test
+    public void test() {
+        int[] ratings = {1, 0, 2};
+        System.out.println(candy(ratings));
+    }
+
+    /**
+     * 思路:需要知道左右分别需要这个位置最少是多少，然后两者中取大，两次贪心
+     * 本质上不算我过得，还有有点印象看过答案
+     * 速度击败98.68%  内存击败58%
+     * @param ratings
+     * @return
+     */
     public int candy(int[] ratings) {
 
+        int[] leftMin = new int[ratings.length];
+
+        leftMin[0] = 1;
+        //记录左边递增序列，每个位置最少需要多少，向右贪心
+        for (int i = 1; i < ratings.length; i++) {
+            if (ratings[i] > ratings[i - 1]) {
+                leftMin[i] = leftMin[i - 1] + 1;
+            }else {
+                leftMin[i]=1;
+            }
+        }
+
+        int result = 0;
+        int lastNeed = 1;
+        //向左贪心，记录每个位置最少需要，加在总和里
+        for (int i = ratings.length - 1; i >= 0; i--) {
+
+            if (i != ratings.length - 1 && ratings[i] > ratings[i + 1]) {
+                lastNeed = lastNeed + 1;
+            } else {
+                lastNeed = 1;
+            }
+            result += Math.max(leftMin[i], lastNeed);
+        }
+
+        return result;
+
     }
 }
diff --git a/Leecode/src/main/java/com/markilue/leecode/greedy/second/T10_LemonadeChange.java b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T10_LemonadeChange.java
new file mode 100644
index 0000000..a067a68
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T10_LemonadeChange.java
@@ -0,0 +1,50 @@
+package com.markilue.leecode.greedy.second;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.greedy.second
+ *@Author: markilue
+ *@CreateTime: 2023-02-10  10:47
+ *@Description:
+ * TODO  力扣860  柠檬水找零:
+ *  在柠檬水摊上，每一杯柠檬水的售价为 5 美元。顾客排队购买你的产品，（按账单 bills 支付的顺序）一次购买一杯。
+ *  每位顾客只买一杯柠檬水，然后向你付 5 美元、10 美元或 20 美元。你必须给每个顾客正确找零，也就是说净交易是每位顾客向你支付 5 美元。
+ *  注意，一开始你手头没有任何零钱。
+ *  给你一个整数数组 bills ，其中 bills[i] 是第 i 位顾客付的账。如果你能给每位顾客正确找零，返回 true ，否则返回 false 。
+ *@Version: 1.0
+ */
+public class T10_LemonadeChange {
+
+    /**
+     * 思路:有大钱先用大钱，没大钱用小钱，因为小钱可以找开所有
+     * @param bills
+     * @return
+     */
+    public boolean lemonadeChange(int[] bills) {
+        int[] count = new int[3];//5,10,20的count
+
+        for (int i = 0; i < bills.length; i++) {
+            if (bills[i] == 5) {
+                count[0]++;
+                continue;
+            }
+
+            if(bills[i]==10){
+                count[0]--;
+                count[1]++;
+
+            }else {
+                //最难讨论的
+                if(count[1]>0){
+                    count[1]--;
+                    count[0]--;
+                }else {
+                    count[0]-=3;
+                }
+            }
+            if(count[0]<0)return false;
+        }
+        return true;
+
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/sort/QuickSort.java b/Leecode/src/main/java/com/markilue/leecode/sort/QuickSort.java
new file mode 100644
index 0000000..9e0f029
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/sort/QuickSort.java
@@ -0,0 +1,69 @@
+package com.markilue.leecode.sort;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.sort
+ *@Author: markilue
+ *@CreateTime: 2023-02-10  12:18
+ *@Description: TODO  自己实现快排:
+ *@Version: 1.0
+ */
+public class QuickSort {
+
+    @Test
+    public void test(){
+        int[] nums={8,5,3,7,6,4,1,9};
+        quickSort(nums,0,nums.length-1);
+        System.out.println(Arrays.toString(nums));
+    }
+
+
+    public void quickSort(int[] nums, int left, int right) {
+        if(left>=right)return;
+        int mid = partition(nums, left, right);
+        quickSort(nums, left, mid - 1);
+        quickSort(nums, mid + 1, right);
+    }
+
+    /**
+     * 寻找num[left]的合适的位置
+     * @param nums
+     * @param left
+     * @param right
+     */
+    public int partition(int[] nums, int left, int right) {
+
+        if (left >= right) return left;
+
+        int start = left + 1;
+        int end = right;
+
+        while (start <= end) {
+            while (start < right && nums[start] < nums[left]) {
+                start++;
+            }
+
+            while (end > left && nums[end] > nums[left]) {
+                end--;
+            }
+            if(start>=end)break;
+            //找到两个需要交换的值
+            swap(nums, start, end);
+            start++;
+            end--;
+
+        }
+        swap(nums, left, end);
+        return end;
+    }
+
+    public void swap(int[] nums, int left, int right) {
+        int temp = nums[left];
+        nums[left] = nums[right];
+        nums[right] = temp;
+    }
+}