# 0438. Find All Anagrams in a String
###### tags: `Leetcode` `Microsoft` `Medium` `Sliding Window` `HashMap`
Link: https://leetcode.com/problems/find-all-anagrams-in-a-string/
## 思路
容易和[0076. Minimum Window Substring](https://hackmd.io/z50NoPGOTlOtJcP9sBE5nw)弄混
这道题是fix length的sliding window 显然要简单很多 不用考虑
**注意第21行的写法，比较两个map，可以直接用equals**
## Code

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> ans = new ArrayList<>();
        Map<Character, Integer> target = new HashMap<>();
        for(int i = 0;i < p.length();i++){
            target.put(p.charAt(i), target.getOrDefault(p.charAt(i), 0)+1);
        }
        Map<Character, Integer> curr = new HashMap<>();
        int len = p.length();
        for(int i = 0;i < s.length();i++){
            curr.put(s.charAt(i), curr.getOrDefault(s.charAt(i),0)+1);
            if(i >= len){
                char ch = s.charAt(i-len);
                if(curr.get(ch)==1){
                    curr.remove(ch);
                }
                else{
                    curr.put(ch, curr.get(ch)-1);
                }
            }
            if(target.equals(curr)){
                ans.add(i-len+1);
            }
        }
        return ans;
    }
}

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        ans = []
        targetCnt = Counter(p)
        currCnt = Counter()
        for i in range(len(s)):
            if i>=len(p):
                currCnt[s[i-len(p)]] -= 1
            currCnt[s[i]] += 1
            if targetCnt==currCnt:
                ans.append(i-len(p)+1)
        return ans