README_EN.md

438. Find All Anagrams in a String

中文文档

Description

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

 

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".

Example 2:

Input: s = "abab", p = "ab"
Output: [0,1,2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

 

Constraints:

• 1 <= s.length, p.length <= 3 * 104
• s and p consist of lowercase English letters.

Solutions

Python3

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        counter = Counter(p)
        ans = []
        left = right = 0
        t = Counter()
        while right < len(s):
            t[s[right]] += 1
            while t[s[right]] > counter[s[right]]:
                t[s[left]] -= 1
                left += 1
            if right - left + 1 == len(p):
                ans.append(left)
            right += 1
        return ans

Java

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int[] counter = new int[26];
        for (char c : p.toCharArray()) {
            ++counter[c - 'a'];
        }
        List<Integer> ans = new ArrayList<>();
        int left = 0, right = 0;
        int[] t = new int[26];
        while (right < s.length()) {
            int i = s.charAt(right) - 'a';
            ++t[i];
            while (t[i] > counter[i]) {
                --t[s.charAt(left) - 'a'];
                ++left;
            }
            if (right - left + 1 == p.length()) {
                ans.add(left);
            }
            ++right;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> counter(26);
        for (char c : p) ++counter[c - 'a'];
        vector<int> ans;
        int left = 0, right = 0;
        vector<int> t(26);
        while (right < s.size())
        {
            int i = s[right] - 'a';
            ++t[i];
            while (t[i] > counter[i])
            {
                --t[s[left] - 'a'];
                ++left;
            }
            if (right - left + 1 == p.size()) ans.push_back(left);
            ++right;
        }
        return ans;
    }
};

Go

func findAnagrams(s string, p string) []int {
	counter := make([]int, 26)
	for _, c := range p {
		counter[c-'a']++
	}
	var ans []int
	left, right := 0, 0
	t := make([]int, 26)
	for right < len(s) {
		i := s[right] - 'a'
		t[i]++
		for t[i] > counter[i] {
			t[s[left]-'a']--
			left++
		}
		if right-left+1 == len(p) {
			ans = append(ans, left)
		}
		right++
	}
	return ans
}

...