Leetcode Distinct Subsequences Java

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.

这道题意思是说，给定两个字符串，S和T，寻找T在S的子序列中出现的次数。

这道题和Eidt Distance类似，都是动态规划的题。

起初想用一个一维数组来做标记数组，但是写起来有一定难度，于是，借鉴了一下别人的思路，然后，在讨论区发现一个理解起来特别简单的方法（https://leetcode.com/discuss/26680/easy-to-understand-dp-in-java）。整理如下

使用一个二维数组dp[T.length+1][S.length+1]做标记。

比如给定两个字符串S：dabdsdsadfsfdes
T：ad
定义一个标记数组：dp[T.length+1][S.length+1];
S：dabdsdsadfsfdes
首先，我们寻找a：dp[1]=[ 0111112222222]
然后，寻找ad: dp[2]=[ 0001122244466]

程序如下：

   public int numDistinct(String S, String T) {
    int[][] mem = new int[T.length()+1][S.length()+1];

    // filling the first row: with 1s
    for(int j=0; j<=S.length(); j++) {
        mem[0][j] = 1;
    }

    // the first column is 0 by default in every other rows but the first, which we need.

    for(int i=0; i<T.length(); i++) {
        for(int j=0; j<S.length(); j++) {
            if(T.charAt(i) == S.charAt(j)) {
                mem[i+1][j+1] = mem[i][j] + mem[i+1][j];
            } else {
                mem[i+1][j+1] = mem[i+1][j];
            }
        }
    }

    return mem[T.length()][S.length()];


    }