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.
int numDistinct(string S, string T) { //1.S[i] != T[j], dp[i][j] = dp[i-1][j], 只能删除 //2.S[i] == T[j], 分两种，保持或删除 dp[i][j] = dp[i-1][j-1] + dp[i-1][j] int len1 = S.size(); int len2 = T.size(); vector<vector<int> >dp(len1+1,vector<int>(len2+1,0)); //初始化 dp[0][0] = 1; for(int i=1; i<=len1;i++) dp[i][0] = 1;//只有一种 for(int i=1; i<=len2;i++) dp[0][i] = 0;//不可能，从1开始 for(int i=1; i<=len1; i++){ for(int j=1; j<=len2; j++){ if(S[i-1] == T[j-1]){ dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; } else{ dp[i][j] = dp[i-1][j]; } } } return dp[len1][len2];}
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