# 0727. Minimum Window Subsequence ###### tags: `Leetcode` `Hard` `Dynamic Programming` Link: https://leetcode.com/problems/minimum-window-subsequence/ ## 思路 ```dp[i][j] => s[1:i]```里最短的、并且包含```t[1:j]```的substring的长度(这个substring必须要求以```s[i]```结尾) 注意边界条件的设定 ```dp[i][0]=0``` ```dp[0][j]=Integer.MAX_VALUE/2``` 因为不可能使得s2是s1的子串 ## Code ```java= class Solution { public String minWindow(String s1, String s2) { int m = s1.length(), n = s2.length(); int[][] dp = new int[m+1][n+1]; for(int i=0; i<=m; i++) Arrays.fill(dp[i], Integer.MAX_VALUE/2); for(int i=0; i<=m; i++) dp[i][0] = 0; for(int i=1; i<=m; i++){ for(int j=1; j<=n; j++){ if(s1.charAt(i-1)==s2.charAt(j-1)){ dp[i][j] = dp[i-1][j-1]+1; } else dp[i][j] = dp[i-1][j]+1; } } int len = Integer.MAX_VALUE/2, idx = -1; for(int i=1; i<=m; i++){ if(len>dp[i][n]){ len = dp[i][n]; idx = i-1; } } return idx==-1?"":s1.substring(idx-len+1, idx+1); } } ```