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：
经典的DP问题，用result[i][j]表示长度为i的S转换到长度为j的T的结果数。
s[i-1]==t[j-1]的时候result[i][j]=result[i-1][j-1]+result[i-1][j];
s[i-1]!=t[j-1]的时候result[i][j]=result[i-1][j];
Acceped Code:
1 class Solution { 2 public: 3 int numDistinct(string s, string t) { 4 if(s.empty()||t.empty()) 5 return 0; 6 if(s.length()<t.length()) 7 return 0; 8 int len1=s.length(); 9 int len2=t.length(); 10 vector<vector<int>> result(len1+1,vector<int>(len2+1,0)); 11 for(int i=0;i<=len1;i++) 12 result[i][0]=1; 13 for(int i=1;i<=len1;i++) 14 for(int j=1;j<=len2;j++) 15 { 16 result[i][j]=result[i-1][j]; 17 if(s[i-1]==t[j-1]) 18 result[i][j]+=result[i-1][j-1]; 19 } 20 return result[len1][len2]; 21 } 22 };