Description

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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.

Solution

跟Edit distance思路很像的DP问题。用dp[i][j]代表s.substring(0, i)中t.substring(0, j)的子序列个数。

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();
        if (m < n) {
            return 0;
        }
        
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                if (j == 0) {
                    dp[i][j] = 1;
                } else if (i < j) {
                    dp[i][j] = 0;
                } else {
                    dp[i][j] = dp[i - 1][j];
                    if (s.charAt(i - 1) == t.charAt(j - 1)) {
                        dp[i][j] += dp[i - 1][j - 1];
                    }
                }
            }
        }
        
        return dp[m][n];
    }
}