leetcode -- 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.

[解题思路]

 DP: let F(i, j) denote the number of ways for S[0]…S[i] contain T[0]..T[j], Then F(n-1, m-1) is our answer and we have
if(S[i] != T[j])  F(i, j) = F(i-1, j);
if(S[i] == T[j]) F(i, j) = F(i-1, j-1) + F(i-1, j);

如果当前字符相同，那有两种可能，用到和不用到这个字符，所以这两种情况加起来。
如果不同，那只能等于用不了这个字符的情况。

public class Solution {
    public int numDistinct(String S, String T) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(S == null || S.length() == 0){
            return 0;
        }
        
        if(T == null || T.length() == 0){
            return 1;
        }
        
        int sLen = S.length(), tLen = T.length();
        if(sLen < tLen){
            return 0;
        }
        
        int[][] dp = new int[sLen + 1][tLen + 1];
        for(int i = 0; i < sLen + 1; i++){
            dp[i][0] = 1;
        }
        
        for(int i = 1; i < tLen + 1; i++){
            dp[0][i] = 0;
        }
        
        for(int i = 1; i < sLen + 1; i++){
            for(int j = 1; j < tLen + 1; j++){
                if(S.charAt(i - 1) == T.charAt(j - 1)){
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        
        return dp[sLen][tLen];
    }
}