LeetCode 567. Permutation in String

原题链接在这里：https://leetcode.com/problems/permutation-in-string/description/

题目：

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].

题解：

如何知道s1是s2某一段的permutation. 只要确定这一段的char count相等就行.

利用sliding window 长度为s1.length累计char count.

Note: check sum == 0 before moving walker. "ab", "bao", runner = 2, walker = 0, check sum == 0.

Time Complexity: O(n). n = s1.length()+s2.length().

Space: O(1).

AC Java:

 1 class Solution {
 2     public boolean checkInclusion(String s1, String s2) {
 3         if(s1 == null || s2 == null || s1.length() > s2.length()){
 4             return false;
 5         }
 6         
 7         int [] map = new int[256];
 8         for(int i = 0; i<s1.length(); i++){
 9             map[s1.charAt(i)]++;
10         }
11         
12         int walker = 0;
13         int runner = 0;
14         int sum = s1.length();
15         while(runner < s2.length()){
16             if(map[s2.charAt(runner++)]-- > 0){
17                 sum--;
18             }
19             
20             if(sum == 0){
21                 return true;
22             }
23             
24             if(runner-walker==s1.length() && map[s2.charAt(walker++)]++ >= 0){
25                 sum++;
26             }
27         }
28         
29         return false;
30     }
31 }