leetcode之Distinct Subsequences

原题如下：

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：这道题看似简单，实则比较复杂，因为首先想到的与人的思维相同的深度遍历的方法行不通，之后借鉴了大牛的解法，还是采用动态规划的思想，而且是二维动态规划的问题，以模式串T为行，以原串S为列构成二维矩阵，动态转移方程分两种情况，如果T[i] !=S[j],则vv[i][j]=vv[i][j-1],否则为vv[i][j]=vv[i][j-1]+vv[i-1][j-1].

class Solution {
public:
    int numDistinct(string S, string T) {
        int m = T.size();
		int n = S.size();
		if(m == 0 || n == 0){
		    return 0;
		}
		vector<vector<int>>vv(m);
		for(int i = 0; i < m; i++){
			vv[i] = vector<int>(n);
		}
		if(S[0] == T[0])
			vv[0][0] = 1;
		else
			vv[0][0] = 0;
		for(int j = 1; j < n; j++){
		    if(S[j] == T[0])
				vv[0][j] = vv[0][j - 1] + 1;
			else
				vv[0][j] = vv[0][j - 1];
		}
		for(int i = 1; i < m; i++){
		    vv[i][0] = 0;
		}
		for(int i = 1; i < m; i++){
			for(int j = 1; j < n; j++){
				vv[i][j] = vv[i][j - 1];
				if(T[i] == S[j]){
				    vv[i][j] +=vv[i - 1][j - 1];
				}
			}
		}
		return vv[m - 1][n - 1];
    }
};