LeetCode115 不同的子序列

最新推荐文章于 2022-04-20 10:18:10 发布

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本文讲解了一种使用动态规划算法解决LeetCode上一道题目“不同子序列”的方法。该问题要求计算字符串S中有多少个不同的子序列等于目标字符串T。通过创建一个二维数组并填充特定值，最终返回数组中特定位置的值即为所求结果。

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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).

Example 1:

Input: S = “rabbbit”, T = “rabbit”
Output: 3
Explanation:

As shown below, there are 3 ways you can generate “rabbit” from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
Example 2:

Input: S = “babgbag”, T = “bag”
Output: 5
Explanation:

As shown below, there are 5 ways you can generate “bag” from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^

https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java

public int numDistinct(String S, String T) {
        // array creation
        int[][] mem = new int[T.length()+1][S.length()+1];

        // filling the first row: with 1s
        for(int j=0; j<=S.length(); j++) {
            mem[0][j] = 1;
        }

        // the first column is 0 by default in every other rows but the first, which we need.

        for(int i=0; i<T.length(); i++) {
            for(int j=0; j<S.length(); j++) {
                if(T.charAt(i) == S.charAt(j)) {
                    mem[i+1][j+1] = mem[i][j] + mem[i+1][j];
                } else {
                    mem[i+1][j+1] = mem[i+1][j];
                }
            }
        }

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