diff --git a/notes/Leetcode 题解 - 贪心思想.md b/notes/Leetcode 题解 - 贪心思想.md
index 13157d86..2e47bfbe 100644
--- a/notes/Leetcode 题解 - 贪心思想.md
+++ b/notes/Leetcode 题解 - 贪心思想.md
@@ -23,7 +23,7 @@
[Leetcode](https://leetcode.com/problems/assign-cookies/description/) / [力扣](https://leetcode-cn.com/problems/assign-cookies/description/)
-```html
+```text
Input: grid[1,3], size[1,2,4]
Output: 2
```
@@ -33,9 +33,9 @@ Output: 2
1. 给一个孩子的饼干应当尽量小并且又能满足该孩子，这样大饼干才能拿来给满足度比较大的孩子。
2. 因为满足度最小的孩子最容易得到满足，所以先满足满足度最小的孩子。
-在以上的解法中，我们只在每次分配饼干时选择一种看起来是当前最优的分配方法，但无法保证这种局部最优的分配方法最后能得到全局最优解。我们假设能得到全局最优解，并使用反证法进行证明，即假设存在一种比我们使用的贪心策略更优的最优策略。如果不存在这种最优策略，表示贪心策略就是最优策略，得到的解也就是全局最优解。
+在以上的解法中，只在每次分配饼干时选择一种看起来是当前最优的分配方法，但无法保证这种局部最优的分配方法最后能得到全局最优解。可以用反证法证明该贪心策略能得到全局最优解：假设存在一种比该贪心策略更优的最优策略。如果不存在这种最优策略，表示贪心策略就是最优策略，得到的解也就是全局最优解。
-证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，可以给该孩子分配第 n 个饼干，并且 m \< n。我们可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
+证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，可以给该孩子分配第 n 个饼干，并且 m \< n。经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
@@ -61,7 +61,7 @@ public int findContentChildren(int[] grid, int[] size) {
[Leetcode](https://leetcode.com/problems/non-overlapping-intervals/description/) / [力扣](https://leetcode-cn.com/problems/non-overlapping-intervals/description/)
-```html
+```text
Input: [ [1,2], [1,2], [1,2] ]
Output: 2
@@ -69,7 +69,7 @@ Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
```
-```html
+```text
Input: [ [1,2], [2,3] ]
Output: 0
@@ -159,7 +159,7 @@ public int findMinArrowShots(int[][] points) {
[Leetcode](https://leetcode.com/problems/queue-reconstruction-by-height/description/) / [力扣](https://leetcode-cn.com/problems/queue-reconstruction-by-height/description/)
-```html
+```text
Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
@@ -214,7 +214,7 @@ public int maxProfit(int[] prices) {
## 6. 买卖股票的最大收益 II
-122\. Best Time to Buy and Sell Stock II (Easy)
+122\. Best Time to Buy and Sell Stock II (Medium)
[Leetcode](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/) / [力扣](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/description/)
@@ -241,7 +241,7 @@ public int maxProfit(int[] prices) {
[Leetcode](https://leetcode.com/problems/can-place-flowers/description/) / [力扣](https://leetcode-cn.com/problems/can-place-flowers/description/)
-```html
+```text
Input: flowerbed = [1,0,0,0,1], n = 1
Output: True
```
@@ -269,11 +269,11 @@ public boolean canPlaceFlowers(int[] flowerbed, int n) {
## 8. 判断是否为子序列
-392\. Is Subsequence (Medium)
+392\. Is Subsequence (Easy)
[Leetcode](https://leetcode.com/problems/is-subsequence/description/) / [力扣](https://leetcode-cn.com/problems/is-subsequence/description/)
-```html
+```text
s = "abc", t = "ahbgdc"
Return true.
```
@@ -293,11 +293,11 @@ public boolean isSubsequence(String s, String t) {
## 9. 修改一个数成为非递减数组
-665\. Non-decreasing Array (Easy)
+665\. Non-decreasing Array (Medium)
[Leetcode](https://leetcode.com/problems/non-decreasing-array/description/) / [力扣](https://leetcode-cn.com/problems/non-decreasing-array/description/)
-```html
+```text
Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
@@ -329,11 +329,11 @@ public boolean checkPossibility(int[] nums) {
## 10. 子数组最大的和
-53\. Maximum Subarray (Easy)
+53\. Maximum Subarray (Medium)
[Leetcode](https://leetcode.com/problems/maximum-subarray/description/) / [力扣](https://leetcode-cn.com/problems/maximum-subarray/description/)
-```html
+```text
For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
```
@@ -359,7 +359,7 @@ public int maxSubArray(int[] nums) {
[Leetcode](https://leetcode.com/problems/partition-labels/description/) / [力扣](https://leetcode-cn.com/problems/partition-labels/description/)
-```html
+```text
Input: S = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
