Distinct Subsequences leetcode 115

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.

题意可以理解将s串通过删除操作变换成t串

动态规划，设置一个矩阵ans[i][j]表示由字符串s前i+1个字符能转换成t串前j+1的种数（ｉ，ｊ从０开始）

１，ｓ［０］＝＝ｔ［０］的话，ａｎｓ［０］［０］＝１，ａｎｓ［ｏ］［ｊ］＝０因为ｓ串长度为１时，并不能转换成ｔ串

２，ｓ［ｉ］＝！ｔ［ｊ］，ａｎｓ［ｉ］［ｊ］＝ａｎｓ［ｉ－１］［ｊ］因为此时字符不相等，该位置能转换成ｔ串的种数就等于ｓ串前ｉ－１位能转换的次数

３，ｓ［ｉ］＝＝ｔ［ｊ］，ａｎｓ［ｉ］［ｊ］＝ａｎｓ［ｉ－１］［ｊ］＋ａｎｓ［ｉ－１］［ｊ－１］，表示此时字符相同，则该位置能转换的次数就等于ｓ串不加该相同字符的种数（ａｎｓ［ｉ－１］［ｊ］ｓ串前ｉ位出现的次数前ｉ－１位也一定会出现）加上ｓ串ｔ串都不包含该字符的种数（ａｎｓ［ｉ－１］［ｊ－１］因为该位字符相同，那么如果ｔ串前ｊ－１位能被ｓ串前ｉ－１位转化，那加上该字符也一定能转化，那么种数就要加上不包含这个字符的种数）

class Solution {public:    int numDistinct(string s, string t) {    if(s.size()==0)return 0;    vector<vector<int>> ans(s.size(),vector<int>(t.size(),0));    if(s[0]==t[0])ans[0][0]=1;    for(int i=1;i<s.size();i++){        for(int j=0;j<t.size();j++){            if(s[i]!=t[j])ans[i][j]=ans[i-1][j];            else ans[i][j]=ans[i-1][j]+(j==0?1:ans[i-1][j-1]);//当j=0时，s[i]==t[0]时，ans[i][j]要加1        }    }    return ans[s.size()-1][t.size()-1];    }};