115. Distinct Subsequences

使用动态规划计算字符串中子序列的数量

最新推荐文章于 2025-09-06 22:33:46 发布

Maggie_2015

最新推荐文章于 2025-09-06 22:33:46 发布

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文章标签：数据结构与算法

本文链接：https://blog.youkuaiyun.com/Maggie_2015/article/details/50630843

本文介绍了一种使用动态规划算法来计算给定字符串中特定子序列的数量的方法。通过实例展示如何将字符串匹配问题转化为动态规划问题，并详细解释了算法背后的逻辑和实现步骤。

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.

动态规划。假设数组table[i][j]代表S的前i个字符同T的前j个字符的Subsequences数。
1、如果S.charAt(i-1) == T.charAt(j-1),那么
table[i][j] += table[i - 1][j] + table[i - 1][j - 1];
2、如果S.charAt(i-1) != T.charAt(j-1),那么
table[i][j] += table[i - 1][j] ;

public int numDistinct(String S, String T) {
        int[][] table = new int[S.length() + 1][T.length() + 1];
        for (int i = 0; i < S.length(); i++)
            table[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)) {
                    table[i][j] += table[i - 1][j] + table[i - 1][j - 1];
                } else {
                    table[i][j] += table[i - 1][j];
                }
            }
        }

        return table[S.length()][T.length()];
    }