438. Find All Anagrams in a String（不太懂）

最新推荐文章于 2025-04-10 01:44:58 发布

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最新推荐文章于 2025-04-10 01:44:58 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/u013573813/article/details/56075168

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本文介绍了一种高效算法，用于找出一个给定字符串中所有指定长度子串的字母异位词起始索引。通过使用哈希表，算法能在O(n)时间内完成任务，适用于仅包含小写英文字母的字符串。

题目

Given a string s and a non-empty string p, find all the start indices ofp's anagrams in s.

Strings consists of lowercase English letters only and the length of both stringss and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:

s: "cbaebabacd" p: "abc"

Output:

[0, 6]

Explanation:

The substring with start index = 0 is "cba", which is an anagram of "abc".

The substring with start index = 6 is "bac", which is an anagram of "abc".

答案解法（还不是很理解）

public class Solution {
    
    
    public List<Integer> findAnagrams(String s, String p) {
    
        List<Integer> list = new ArrayList<>();
        if (s == null || s.length() == 0 || p == null || p.length() == 0) return list;
        int[] hash = new int[256]; 
        for (char c : p.toCharArray()) 
            hash[c]++;
        int left = 0, right = 0, count = p.length();
        while (right < s.length()) {
            if (hash[s.charAt(right)] >= 1)
              count--;
            hash[s.charAt(right)]--;
            right++;
            if (count == 0) 
                list.add(left);
            if (right - left == p.length() ) {
                if (hash[s.charAt(left)] >= 0)
                    count++;
                hash[s.charAt(left)]++;
                left++;
            }
        }
        return list;
    }
}

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