[LeetCode] 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.

思路：用递归超时

DP： 建立一个matrix，matrix（i，j）表示从s.substr(0,j) 可以有多少个t.substr(0,i)的子串

<pre name="code" class="cpp">class Solution {public:    int numDistinct(string s, string t) {        int m = s.length();        int n = t.length();        if(m*n==0) return 0;                // each position<i,j> indicates how many subsequences of t.substr(0,i) contains in s.substr(0,j)        vector<vector<int>> subseqs;        //init it to all zeros        for(int i=0; i<n; i++)        {            vector<int> row;            for(int j=0; j<m; j++)            {                row.push_back(0);            }            subseqs.push_back(row);        }                //int the first row        for(int j=0; j<m; j++)        {            if(j==0){                if(t[0]==s[j]) subseqs[0][j]=1;                continue;            }            if(t[0]!=s[j]) subseqs[0][j]=subseqs[0][j-1];            else subseqs[0][j]=subseqs[0][j-1]+1;        }                for(int i=1; i<n; i++)        {            for(int j=1; j<m; j++)            {                if(t[i]!=s[j]){                    subseqs[i][j] = subseqs[i][j-1];                }                else{                    // the first term is: if s[j] is not used to form the substr t.substr(0,j) from s.substr(0,i)                    // the second term is s[j] is used to get the subsequence                    subseqs[i][j] = subseqs[i][j-1]+subseqs[i-1][j-1];                }            }        }        return subseqs[n-1][m-1];    }};