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. 

分析：此题题目叙述比较不明确，改为"count the number of distinct subsequence T in S"， 即找到S中等于T的subsequence个数。我们可以用动态规划的方法来解。我们用record[i][j]表示S[0:i-1]中等于T[0:j-1]的subsequence个数。那么当S[i-1] == T[j-1]时，record[i][j] = record[i-1][j] + record[i-1][j-1]；当S[i-1] != T[j-1]时，record[i][j] = record[i-1][j]。在实现时，可以用一个二维数组保存record，也可以用一个一维数组，二维数组的方式更容易理解。

1.　二维数组实现：

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

2.　一维数组实现，我们可以用两个长度为T.length()+1的一维数组prev和cur，prev记录record[i-1][j]， cur记录record[i][j]，因为记录关于i的record时，只用到i-1的信息。其实我们可以更节省空间，只用一个一维数组，但要从后往前遍历j，因为要保证当前被更新的旧值不会被以后的计算用到。代码如下：

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

Leetcode: Distinct Subsequences