[leetcode]Distinct Subsequences

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. 

算法思路：

思路1：递归，求出所有长度为t.length()的子串，相比会超时

思路2：DP；dp[i][j]代表从s[0..i]中删除s.length - t.length个字符得到t[0..j]的不同删除方法数。

• 初始化：如果t为空，任意的s删除到t都是1种。
• 如果S[i] = T[j], dp[i][j] = dp[i-1][j-1]+dp[i][j - 1]
• 如果S[i] 不等于 T[j], dp[i][j] = dp[i][j - 1]

代码如下：

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