118 · 不同的子序列Distinct Subsequences

CMYing85

于 2021-06-11 16:05:51 发布

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分类专栏：算法文章标签：算法动态规划字符串

本文链接：https://blog.youkuaiyun.com/caomengying/article/details/117820477

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这篇博客介绍了一种计算字符串S中等于目标字符串T的不同子序列数量的方法。通过动态规划实现，遍历S和T，当字符匹配时，当前子序列计数增加，最后返回dp矩阵的最后一个元素作为结果。例如，对于S='rabbbit'和T='rabbit'，输出为3。

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118 · 不同的子序列Distinct Subsequences

描述
Given two strings S and 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)
Input:

S = “rabbbit”
T = “rabbit”
Output:

3
Explanation:

You could remove any ‘b’ in S, so there are 3 ways to get T.

public class Solution {
    /**
     * @param S: A string
     * @param T: A string
     * @return: Count the number of distinct subsequences
     */
    public int numDistinct(String S, String T) {
        // write your code here
        int n = S.length() , m = T.length() ;
        int[][] dp = new int[m+1][n+1] ;
        for(int i = 0 ; i <= n ; i++){
            dp[0][i] = 1 ;
        }
        for(int i = 1 ; i <= m ; i++){
            for(int j = 1 ; j <= n ; j++){
                 if(T.charAt(i-1)==S.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1] +dp[i][j-1];
                else
                    dp[i][j] = dp[i][j-1];

            }
        }
        return dp[dp.length-1][dp[0].length-1];
    }
}