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.

一刷
用dp来求解。构造二维数组dp[s.length()+1][t.length()+1], dp[i][0]=1;(当t为[]时)，如果s.charAt(i-1) == t.charAt(j-1), dp[i][j] = dp[i-1][j-1] + dp[i-1][j], 如果不等， dp[i][j] = dp[i-1][j]

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

二刷
空间复杂度O(m)
由于我们仅用到了dp[i-1][j-1], dp[i-1][j]，考虑用一维数组替换二维数组
对于i, j， 未修改dp[j]时，此时dp[j]表示dp[i-1][j]， 对于下一个j, 表示为dp[i-1][j-1]

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