[LintCode] 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).

Example

Given S = "rabbbit", T = "rabbit", return 3.

Challenge

Do it in O(n2) time and O(n) memory.

O(n2) memory is also acceptable if you do not know how to optimize memory.

 

SOLUTION: 

以下大部分摘自 http://www.cnblogs.com/yuzhangcmu/p/4196373.html 

遇到这种两个串的问题，很容易想到DP。但是这道题的递推关系不明显。可以先尝试做一个二维的表int[][] dp，用来记录匹配子序列的个数（以S ="rabbbit",T = "rabbit"为例）：

 

    r a b b b i t

  1 1 1 1 1 1 1 1

r 0 1 1 1 1 1 1 1

a 0 0 1 1 1 1 1 1

b 0 0 0 1 2 3 3 3

b 0 0 0 0 1 3 3 3

i 0 0 0 0 0 0 3 3

t 0 0 0 0 0 0 0 3  

 

看这个图更直接，我们看如果S(i) == T(j) 时，先看S(i - 1)跟T(j - 1) 去匹配，然后还有一种情况就是用S(i - 1)跟T(j) 去匹配 ==> result[i][j] = result[i - 1][j - 1] + result[i - 1][j]==>意思就是既然这个位置两个位置相等了，那么T(j)可以匹配也可以不匹配，有两个情况，都要考虑。

如果S(i) != T(j) 那么就必须看S(i - 1) T(j - 1)之间能不能匹配了 ==> result[i][j] = result[i - 1][j - 1]

再看初始化，第一行第一列明显需要初始化，第一行意思是，S不空，T是空的，所以肯定存在1个（空集），第一列就是说，S空，不管T是什么，肯定是0个LCS。

看代码：

public class Solution {
    /**
     * @param S, T: Two string.
     * @return: Count the number of distinct subsequences
     */
    public int numDistinct(String S, String T) {
        if (S == null || T == null){
            return 0;
        }
        int[][] result = new int[S.length() + 1][T.length() + 1];
        for (int i = 0; i <= S.length(); i++){
            result[i][0] = 1;
        }
        for (int i = 1; i <= S.length(); i++){
            for (int j = 1; j <= T.length(); j++){
                if (S.charAt(i - 1) == T.charAt(j - 1)){
                    result[i][j] = result[i - 1][j] + result[i - 1][j - 1];//needs T[j] or not
                } else {
                    result[i][j] = result[i - 1][j];
                }
            }
        }
        return result[S.length()][T.length()];
    }
}