438. Find All Anagrams in a String - Easy

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".

 

用sliding window模板，保存结果的判断条件是fast - slow == p.length()

time: O(n), space: O(m)　　-- n: length of s, m: length of p

class Solution {    
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        if(s.length() == 0 || s.length() < p.length()) {
            return res;
        }
        
        Map<Character, Integer> map = new HashMap<>();
        for(char c : p.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        
        int counter = map.size(), slow = 0, fast = 0;
        
        while(fast < s.length()) {
            char c = s.charAt(fast);
            if(map.containsKey(c)) {
                map.put(c, map.get(c) - 1);
                if(map.get(c) == 0) {
                    counter--;
                }
            }
            fast++;
            
            while(counter == 0) {
                char tempc = s.charAt(slow);
                if(map.containsKey(tempc)) {
                    map.put(tempc, map.get(tempc) + 1);
                    if(map.get(tempc) > 0) {
                        counter++;
                    }
                }
                if(fast - slow == p.length()) {
                    res.add(slow);
                }
                slow++;
            }
        }
        return res;
    }
}

 

二刷：

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        Map<Character, Integer> map = new HashMap<>();
        for(char c : p.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        
        int slow = 0, fast = 0, counter = map.size();
        while(fast < s.length()) {
            char c = s.charAt(fast);
            
            if(map.containsKey(c)) {
                int cnt = map.get(c);
                map.put(c, cnt - 1);
                if(cnt == 1) {
                    counter--;
                }
            }
            fast++;
            
            while(counter == 0) {
                if(fast - slow == p.length()) {
                    res.add(slow);
                }
                char t = s.charAt(slow);
                if(map.containsKey(t)) {
                    int cnt2 = map.get(t);
                    map.put(t, cnt2 + 1);
                    if(cnt2 == 0) {
                        counter++;
                    }
                }
                slow++;
            }
        }
        return res;
    }
}