lcha

好厉害，我要看好一会才能理解，大佬！

712 两个字符串最小ASCII删除和 自底向上dp解法 ```java public int minimumDeleteSum(String s1, String s2) { int m = s1.length(), n = s2.length(); //s1[...i]和s2[...j]若要成为相等的字符串所需要删除的字符的ASCII值的最小和为dp[i][j] int[][] dp = new int[m + 1][n + 1]; //base case...

这里的reverse使用前边的递归反转实现 ```java public boolean isPalindrome(ListNode head) { ListNode slow, fast; slow = fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next;...

备忘录+递归确实会超时 在第74个用例的时候,输入数据是: s = 6 , n = 10000 ``` class Solution { int[][] memo; public int superEggDrop(int k, int n) { memo = new int[k + 1][n + 1];...