[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.

动态规划。将“S串中前m个字母的子串中包含多少个T串中前n个字母的子串”这样的问题记为A[m][n]。 可以得到递推式 ：
if(S[m-1] == T[n-1]) A[m][n] = A[m-1][n-1] + A[m-1][n];
else A[m][n] = A[m-1][n-1];
再处理边界情况即可。简单起见，用类似打表记录式的递归实现。

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int **a;
        a = new int*[S.length()];
        for(int i = 0;i < S.length();i++)
            a[i] = new int[T.length()];
        for(int i = 0;i < S.length();i++)
            for(int j = 0;j < T.length();j++)
                a[i][j] = -1;
        return DP(S.length()-1,T.length()-1,a,&S,&T);
    }
    int DP(int m,int n,int **a,string *S, string *T)
    {
        if(n == -1)
            return 1;
        else if(m == -1)
            return 0;
        if(m < n)
            return 0;
        if(a[m][n] != -1)
            return a[m][n];
        if(S->at(m) != T->at(n))
        {
            a[m][n] = DP(m-1,n,a,S,T);
            return a[m][n];
        }
        else
        {
            a[m][n] = DP(m-1,n-1,a,S,T) + DP(m-1,n,a,S,T);
            return a[m][n];
        }
    }
};