Distinct Subsequences

最新推荐文章于 2021-03-20 14:23:47 发布

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最新推荐文章于 2021-03-20 14:23:47 发布

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分类专栏： Dynamic Programming

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/41842317

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本文介绍了一种使用动态规划解决字符串子序列匹配问题的方法，通过构建矩阵来追踪两个字符串间可能的子序列数量。

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.

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Have you met this question in a real interview?

这道题首先引用我忘记在哪里看到的一句话：

“When you see string problem that is about subsequence or matching, dynamic programming method should come to your mind naturally. ”

所以这种类型题可以多往DP思考思考。

思路不是很明显的时候，可以画图表。

画图之后，可以观察，得到规律：

左边是S,上面是 T.

matrix[i][j] 表示：S在i, T在j 的时候，S能够表示T的种数。

if S[i] != T[j] 的时候， matrix[i][j] = matrix[i-1][j] 也就是说，like: S abc T a, 当move到b的时候，b的个数就等于a which is 1, c的个数也等于a which is 1.

if S[i] == T[j] , matrix[i][j] = matrix[i-1][j] + matrix[i-1][j-1].这点是从图中

2 1

推理出来的。可以理解为： S[i] == T[j] 了，数目可以由原来的，各自倒退一步，S[i-1] T[j-1]的数目， + matrix[i-1][j]

因为matrix[i][j] 的数目，至少是matrix[i-1][j] 这个看矩阵第二列，也就是T=r那一列。全部是：

S = rabbit T = r

递推的base：

S="", T 这一行全部是0

T="", S 这一列全部是1，因为""为任何string的sequence，just delete this string will get ""

看不懂就看看这个： http://blog.youkuaiyun.com/abcbc/article/details/8978146

public class Solution {
    public int numDistinct(String S, String T) {
        if(S == null || T == null || T.length()>S.length()){return 0;}
        int[][] matrix = new int[S.length()+1][T.length()+1];
        
        // "" is 1 sequence of "";
        matrix[0][0] = 1;
        
        //init 1st col; T is "";
        for(int i=1; i<S.length()+1; i++){
            matrix[i][0] = 1;
        }
        
        //init 1st row; S is "";
        for(int i=1; i<T.length()+1; i++){
            matrix[0][i] = 0;
        }
        
        for(int i=1; i<S.length()+1; i++){
            for(int j=1; j<T.length()+1; j++){
                if(S.charAt(i-1) == T.charAt(j-1)){
                    matrix[i][j] = matrix[i-1][j]+matrix[i-1][j-1];
                }else{
                    matrix[i][j] = matrix[i-1][j];
                }
            }
        }
        return matrix[S.length()][T.length()];
    }
}