premutation in string medium

Author: weiy

String/PermutationInString.py

@@ -0,0 +1,74 @@
+"""
+Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
+Example 1:
+Input:s1 = "ab" s2 = "eidbaooo"
+Output:True
+Explanation: s2 contains one permutation of s1 ("ba").
+Example 2:
+Input:s1= "ab" s2 = "eidboaoo"
+Output: False
+Note:
+The input strings only contain lower case letters.
+The length of both given strings is in range [1, 10,000].
+
+类似于 Find All Anagrams in a String 难度应该颠倒过来。
+
+这个的测试用例更丰富，发现了没想到的一个盲点。
+
+思路请看 https://github.com/HuberTRoy/leetCode/blob/master/DP/FindAllAnagramsInAString.py
+
+
+beat 79%
+
+测试地址：
+https://leetcode.com/problems/permutation-in-string/description/
+
+
+"""
+class Solution(object):
+    def checkInclusion(self, s1, s2):
+        """
+        :type s1: str
+        :type s2: str
+        :rtype: bool
+        """
+        if len(s1) > len(s2):
+            return False
+        
+        counts = {}
+        
+        for i in s1:
+            try:
+                counts[i] += 1
+            except:
+                counts[i] = 1
+        
+        pre = counts.copy()
+        
+        for c in range(len(s2)):
+            i = s2[c]
+            if i in pre:
+                pre[i] -= 1
+                if not pre[i]:
+                    pre.pop(i)
+                
+                if not pre:
+                    return True
+            else:
+                if i in counts:
+                    if i != s2[c-len(s1)+sum(pre.values())]:
+                        for t in s2[c-len(s1)+sum(pre.values()):c]:
+                            if t == i:
+                                break
+                            try:
+                                pre[t] += 1
+                            except:
+                                pre[t] = 1
+                    continue
+                pre = counts.copy()
+                if i in pre:
+                    pre[i] -= 1
+                    if not pre[i]:
+                        pre.pop(i)
+        
+        return False