Leetcode习题：Distinct Subsequences

最新推荐文章于 2019-05-06 23:39:39 发布

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本文探讨了一种高效方法来计算一个字符串中特定子序列出现的次数，利用递归公式和动态规划技巧解决字符串匹配问题。

Distinct Subsequences

Total Accepted: 2790 Total Submissions: 12196 My Submissions

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.

主要在于找出递推关系式：

N[i,j] = N[i-1,j] + if(S[i] == T[j])? N[i-1, j -1]:0;

代码就可以写出：

public class Solution {
    public int numDistinct(String S, String T) {
        int[][] num = new int[S.length()+1][T.length()+1];
        for(int i = 0; i<=S.length(); i++){
            num[i][0] = 0;
        }
        for(int j=0; j<=T.length(); j++){
            num[0][j] = 0;
        }
        
        for(int i = 1; i<=S.length(); i++){
            for(int j = 1; j<=i && j<=T.length(); j++){
                num[i][j] = num[i-1][j];
                if(S.charAt(i-1) == T.charAt(j-1)){
                    if(j == 1){
                        num[i][j] += 1;
                    }else{
                        num[i][j] += num[i-1][j-1];
                    }
                }
            }
        }
        return num[S.length()][T.length()];
    }
}