438. Find All Anagrams in a String - jiejackyzhang/leetcode-note GitHub Wiki

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

Strings consists of lowercase English letters only and the length of both strings s 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".

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".

解题思路为sliding window。

采用一个数组记录p中的字符及其次数。 使用两个指针left和right作为sliding window来扫描s。

public class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        if(s == null || s.length() == 0 || p == null || p.length() == 0) return res;
        int[] map = new int[26];
        for(int i = 0; i < p.length(); i++) {
            map[p.charAt(i)-'a']++;
        }
        int left = 0, right = 0;
        int count = p.length();
        while(right < s.length()) {
            char r = s.charAt(right);
            //move right everytime, if the character exists in p's hash, decrease the count
            //current hash value >= 1 means the character is existing in p
            if(map[r-'a'] >= 1) count--;
            map[r-'a']--;
            right++;
            //when the count is down to 0, means we found the right anagram
            //then add window's left to result list
            if(count == 0) res.add(left);
            //if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window
            //++ to reset the hash because we kicked out the left
            //only increase the count if the character is in p
            //the count >= 0 indicate it was original in the hash, cuz it won't go below 0
            if(right - left == p.length()) {
                char l = s.charAt(left);
                if(map[l-'a'] >= 0) count++;
                map[l-'a']++;
                left++;
            }
        }
        return res;
    }
}
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