LeetCode OJ

题目：

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.

解题思路：

使用动态规划：设dp[i][j]表示S的前i个字符和T的前j个字符中的解。

 if (S[i - 1] == T[j - 1]) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];

 else dp[i][j] = dp[i-1][j];

代码：

class Solution {
public:
    int numDistinct(string S, string T) {
        int rows = S.length(), cols = T.length();
        vector<vector<int>> dp(rows + 1, vector<int>(cols + 1, 0));
        for (int i = 0; i < rows; i++) {
            dp[i][0] = 1; // if T is empty, every S has 1 match
        }

        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                if (S[i - 1] == T[j - 1]) {
                    // have two situation
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }
                else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[rows][cols];
    }
};