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Update Leetcode 题解 - 贪心思想.md
代码经力扣判题验证；难度标注按力扣更新，第一人称改为客观陈述，示例代码块的语言标注改为 text。
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CyC2018
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[Leetcode](https://leetcode.com/problems/assign-cookies/description/) / [力扣](https://leetcode-cn.com/problems/assign-cookies/description/)
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```html
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```text
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Input: grid[1,3], size[1,2,4]
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Output: 2
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```
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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。
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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。
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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。
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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。
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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。
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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。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/e69537d2-a016-4676-b169-9ea17eeb9037.gif" width="430px"> </div><br>
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@@ -61,7 +61,7 @@ public int findContentChildren(int[] grid, int[] size) {
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[Leetcode](https://leetcode.com/problems/non-overlapping-intervals/description/) / [力扣](https://leetcode-cn.com/problems/non-overlapping-intervals/description/)
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```html
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```text
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Input: [ [1,2], [1,2], [1,2] ]
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Output: 2
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Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
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```
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```html
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```text
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Input: [ [1,2], [2,3] ]
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Output: 0
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@@ -159,7 +159,7 @@ public int findMinArrowShots(int[][] points) {
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[Leetcode](https://leetcode.com/problems/queue-reconstruction-by-height/description/) / [力扣](https://leetcode-cn.com/problems/queue-reconstruction-by-height/description/)
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```html
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```text
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Input:
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[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
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## 6. 买卖股票的最大收益 II
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122\. Best Time to Buy and Sell Stock II (Easy)
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122\. Best Time to Buy and Sell Stock II (Medium)
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[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/)
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@@ -241,7 +241,7 @@ public int maxProfit(int[] prices) {
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[Leetcode](https://leetcode.com/problems/can-place-flowers/description/) / [力扣](https://leetcode-cn.com/problems/can-place-flowers/description/)
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```html
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```text
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Input: flowerbed = [1,0,0,0,1], n = 1
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Output: True
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```
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@@ -269,11 +269,11 @@ public boolean canPlaceFlowers(int[] flowerbed, int n) {
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## 8. 判断是否为子序列
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392\. Is Subsequence (Medium)
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392\. Is Subsequence (Easy)
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[Leetcode](https://leetcode.com/problems/is-subsequence/description/) / [力扣](https://leetcode-cn.com/problems/is-subsequence/description/)
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```html
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```text
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s = "abc", t = "ahbgdc"
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Return true.
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```
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## 9. 修改一个数成为非递减数组
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665\. Non-decreasing Array (Easy)
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665\. Non-decreasing Array (Medium)
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[Leetcode](https://leetcode.com/problems/non-decreasing-array/description/) / [力扣](https://leetcode-cn.com/problems/non-decreasing-array/description/)
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```html
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```text
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Input: [4,2,3]
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Output: True
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Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
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@@ -329,11 +329,11 @@ public boolean checkPossibility(int[] nums) {
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## 10. 子数组最大的和
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53\. Maximum Subarray (Easy)
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53\. Maximum Subarray (Medium)
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[Leetcode](https://leetcode.com/problems/maximum-subarray/description/) / [力扣](https://leetcode-cn.com/problems/maximum-subarray/description/)
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```html
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```text
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For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
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the contiguous subarray [4,-1,2,1] has the largest sum = 6.
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```
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@@ -359,7 +359,7 @@ public int maxSubArray(int[] nums) {
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[Leetcode](https://leetcode.com/problems/partition-labels/description/) / [力扣](https://leetcode-cn.com/problems/partition-labels/description/)
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```html
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```text
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Input: S = "ababcbacadefegdehijhklij"
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Output: [9,7,8]
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Explanation:
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