From c2e7444265d97287ad79aecd56610e4d303154f6 Mon Sep 17 00:00:00 2001
From: KEQI HUANG <keqih@smu.edu>
Date: Fri, 20 Oct 2017 16:57:33 -0500
Subject: [PATCH] Create 438._Find_All_Anagrams_in_a_String.md

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+### 438. Find All Anagrams in a String
+
+
+题目:
+<https://leetcode.com/problems/Find-All-Anagrams-in-a-String/>
+
+
+难度:
+
+Easy
+
+
+
+思路
+
+刚开始打算直接遍历整个s，时间复杂度为O(m*n),m和n分别为字符串p和s的长度，但是超时了
+
+
+
+```python
+python
+class Solution(object):
+    def findAnagrams(self, s, p):
+        """
+        :type s: str
+        :type p: str
+        :rtype: List[int]
+        """
+        l, res = len(p), []
+        for i in range(len(s)):
+            if collections.Counter(s[i:i+l]) == collections.Counter(p):
+                res.append(i)
+        return res
+```
+于是用双指针，left和right都从0开始往后遍历
+```python
+class Solution(object):
+    def findAnagrams(self, s, p):
+        """
+        :type s: str
+        :type p: str
+        :rtype: List[int]
+        """
+        result = []
+
+        cnts = [0] * 26
+        for c in p:
+            cnts[ord(c) - ord('a')] += 1
+        
+        left, right = 0, 0
+        while right < len(s):
+            cnts[ord(s[right]) - ord('a')] -= 1
+            while left <= right and cnts[ord(s[right]) - ord('a')] < 0:
+                cnts[ord(s[left]) - ord('a')] += 1
+                left += 1
+            if right - left + 1 == len(p):
+                result.append(left)
+            right += 1
+
+        return result
+```
+
+
+
+Author: Keqi Huang
+
+If you like it, please spread your support
+
+![Support](https://github.com/Lisanaaa/myTODOs/blob/master/WechatIMG17.jpeg)
-- 
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