115. Distinct Subsequences (String; DP)

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：

dp[i][j]表示 # of T[0...j-1] in S[0...i-1] (dp[0][0]表示s=NULL，t=NULL的情况）

如果S[i]！=T[j]，那么dp[i][j]=dp[i-1][j]

如果S[i]=T[j]，dp[i][j]=dp[i-1][j]+j抽出的情况=dp[i-1][j]+dp[i-1][j-1] (注意：这里并不是简单的dp[i-1][j]+1, j抽出后，dp[i-1][j-1]是要大于dp[i-1][j]的)

class Solution {
public:
    int numDistinct(string s, string t) {
        int sLen = s.length();
        int tLen = t.length();
        vector<vector<int>> dp(sLen+1, vector<int>(tLen+1,0));
        for(int i = 0; i <= sLen; i++){ //if t==NULL, 1 method to match
            dp[i][0]=1;
        }
        
        for(int i = 1; i <=sLen; i++){
            for(int j = 1; j <= tLen; j++){
                if(s[i-1]==t[j-1]){
                    dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
                }
                else{
                    dp[i][j]=dp[i-1][j];
                }
            }
        }
        return dp[sLen][tLen];
    }
};