567. Permutation in String

最新推荐文章于 2019-05-15 16:49:38 发布

mingyangdai

最新推荐文章于 2019-05-15 16:49:38 发布

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分类专栏：数据结构&算法文章标签：算法

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本文介绍了一种用于检查一个字符串是否包含另一个字符串的所有排列的方法。通过使用整数数组而非哈希表来跟踪字符出现次数，该算法能高效地解决LeetCode上的“字符串中的排列”问题。

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Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.

Example 1:

Input:s1 = "ab" s2 = "eidbaooo"
Output:True
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:

Input:s1= "ab" s2 = "eidboaoo"
Output: False

第二遍刷做出来了用的map 答案使用的数组比较简洁

public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length())
            return false;
        int[] s1map = new int[26];
        int[] s2map = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            s1map[s1.charAt(i) - 'a']++;
            s2map[s2.charAt(i) - 'a']++;
        }
        for (int i = 0; i < s2.length() - s1.length(); i++) {
            if (matches(s1map, s2map))
                return true;
            s2map[s2.charAt(i + s1.length()) - 'a']++;
            s2map[s2.charAt(i) - 'a']--;
        }
        return matches(s1map, s2map);
    }
    public boolean matches(int[] s1map, int[] s2map) {
        for (int i = 0; i < 26; i++) {
            if (s1map[i] != s2map[i])
                return false;
        }
        return true;
    }

还有优化空间

https://leetcode.com/problems/permutation-in-string/solution/