Leetcode 727. Minimum Window Subsequence

Problem:

Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W.

If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.

Example 1:

Input: 
S = "abcdebdde", T = "bde"
Output: "bcde"
Explanation: 
"bcde" is the answer because it occurs before "bdde" which has the same length.
"deb" is not a smaller window because the elements of T in the window must occur in order.

 

Note:

• All the strings in the input will only contain lowercase letters.
• The length of S will be in the range [1, 20000].
• The length of T will be in the range [1, 100].

Solution:

　　这道题拿到手的第一想法应该是动态规划，用dp[i][j]代表前i个T中的字符和前j个S中的字符在S中满足条件的最大开始位置，说起来很绕，以Example 1为例，首先初始化第一行，当碰T[i]等于S[j]时，将j赋值给dp[i][j]，否则dp[i][j]等于dp[i][j-1]。对于其余的行，如果T[i]等于S[j]，那么dp[i][j]等于dp[i-1][j]，为什么要这么做？那是因为我知道了前i-1个T字符串和前j个S字符串在S中的最大开始位置，而T[i]和S[j]又是相等的，那么即使多了一个T中的i位置的字符也不会影响结果，所以dp[i][j]=dp[i-1][j]，如果T[i]和S[j]不等，那么dp[i][j]等于dp[i][j-1]。但这里有个陷阱，看"cnhczmccqouqadqtmjjzl"和"mm"这个例子，如果简单的按照这个逻辑，T的第二个m会和S的第一个m重复使用，因此需要多一布判断T[i]和T[i-1]是否相同，如果相同的话dp[i][j]就等于dp[i-1][j-1].

	a	b	c	d	e	b	d	d	e
b	-1	1	1	1	1	5	5	5	5
d	-1	-1	-1	1	1	1	5	5	5
e	-1	-1	-1	-1	1	1	1	1	5

 

Code:

 

class Solution {
public:
    string minWindow(string S, string T) {
        int m = S.size();
        int n = T.size();
        vector<vector<int>> dp(n,vector<int>(m,-1));
        for(int j = 0;j != m;++j){
            if(T[0] == S[j])
                dp[0][j] = j;
            else{
                if(j != 0)
                    dp[0][j] = dp[0][j-1];
            }
        }
        for(int i = 1;i != n;++i){
            for(int j = 0;j != m;++j){
                if(S[j] == T[i]){
                    if(T[i] == T[i-1]){
                        if(j != 0)
                            dp[i][j] = dp[i-1][j-1];
                    }
                    else
                        dp[i][j] = dp[i-1][j];
                }
                else{
                    if(j != 0)
                        dp[i][j] = dp[i][j-1];
                }
            }
        }
        string result = "";
        int minlength = INT_MAX;
        for(int j = 0;j != m;++j){
            if(dp[n-1][j] != -1 && j-dp[n-1][j]+1 < minlength){
                minlength = j-dp[n-1][j]+1;
                result = S.substr(dp[n-1][j],minlength);
            }
        }
        return result;
    }
};