【leetcode】No.115：distinct-subseq

题目描述

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie,"ACE"is a subsequence of"ABCDE"while"AEC"is not).
Here is an example:
S ="rabbbit", T ="rabbit"
Return 3.

思路

动态规划问题，设dp[i][j]表示S.subsequence(0, i)中不重复的T.subsequence(0, j)的个数，对于T来说，如果T[i] != S[i]，则dp[i][j] = dp[i-1][j]，加入第i个字符没有增益，如果T[i] == S[i]，则dp[i][j]的来源有两个，j字符加入或者j字符不加入，这两种情况的个数都能原封不动的继承，所以dp[i][j] = dp[i-1][j-1] + dp[i-1][j]

代码

public class Solution {
    public int numDistinct(String S, String T) {
        int sLen = S.length();
        int tLen = T.length();
        int dp[] = new int[tLen+1];
        dp[0] = 1;
        for (int i=0; i<sLen; i++){
            for (int j=tLen; j>0; j--){
                if (S.charAt(i) == T.charAt(j-1)){
                    dp[j] = dp[j] + dp[j-1];
                }
            }
        }
        return dp[tLen];
    }
}