115.Distinct Subsequences

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.

这一题采用的是dp的思路。设置dp[i][j]表示字符串t[0->j)在字符串s[0->i)中出现的的数目。初始化设置dp[i][0] i∈[0,s.length() ),表示字符串s[0->i)变为空字符串只有一种方法。对于i∈[1,s.length()] j∈[1,t.length()],如果s[i-1]==t[j-1],那么dp[i][j]=dp[i-1][j-1]+dp[i-1][j],表示取s[i-1]对应t[j-1],或者不取s[i-1]对应t[j-1]的所有子串之和。如果如果s[i-1]！=t[j-1]，那么不能用s[i-1]对应t[j-1],dp[i][j]只能是dp[i-1][j]。

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