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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：求S中所有子串中与T相同的有多少个
空串是任意字符串的子串。
代码：
class Solution { public: int numDistinct(string S, string T) { int row=S.size()+1; int col=T.size()+1; vector<int> temp(col,0); vector<vector<int>> dp(row,temp); dp[0][0]=1; for(int i=1;i<row;++i) dp[i][0]=1; for(int i=1;i<row;++i){ for(int j=1;j<=i&&j<col;++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[row-1][col-1]; } };
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原文地址：http://www.cnblogs.com/fightformylife/p/4242233.html
