[LeetCode] 115. Distinct Subsequences Java

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题目:

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.

题意及分析:给定字符串S和T,若S中删除某些字符之后能得到T,求有几种方式。使用动态规划的方式求解,使用一个数组men[s.length+1][t.length+1]来保存S从开始位置到i和T从开始位置到j能够匹配的个数,即S(0,i)中包含多少个T(0,j)的子序列。

这里分三种情况:

1、i==0,即s为空,那么men[0][j]=0

2、j==0,即t为空,那么men[i][0]=1

3、对于i>0&&j>0有:

 (1)若s.charAt(i-1) !=t.chatAt(j-1),那么s的i点加入不加入影响都是一样的,即men[i][j] = men[i-1][j]

 (2)若s.charAt(i-1) ==t.chatAt(j-1),那么当前字符可选择匹配或者是不匹配,所以men[i][j] = men[i - 1][j -1] + men[i - 1][j];

代码:

class Solution {
    public int numDistinct(String s, String t) {
        int[][] men = new int[s.length()+1][t.length()+1];      //men[i][j]代表s从开始位置到i,t从开始位置到j 子串匹配的个数,可以得到men[i][j]>=men[i-1][j]

        //当i为0,即s子串为空的时候,men[i][j]==0
        for(int j=0;j<t.length();j++){
            men[0][j] =0;
        }
        //当j=0,即t子串为空,men[i][j]=1
        for (int i=0;i<s.length();i++){
            men[i][0] = 1;
        }
        for(int i=1;i<=s.length();i++){
            for(int j=1;j<=t.length();j++){
                men[i][j] = men[i-1][j];
                if(s.charAt(i-1)==t.charAt(j-1)){
                    men[i][j] += men[i-1][j-1];
                }
            }
        }
        return men[s.length()][t.length()];
    }
}

 


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