438. Find All Anagrams in a Strin

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".

本题开始的时候并没有想用O(n)的时间复杂度做出来,而是用了O(mn)的时间复杂度做,看了答案才知道本题是sliding window的变体,是two pointer的例子,不是很难,代码如下:
 1 public class Solution {
 2     public List<Integer> findAnagrams(String s, String p) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if(s==null||s.length()==0||p==null||p.length()==0) return res;
 5         int left = 0,right = 0;
 6         int count  =p.length();
 7         int[] word = new int[256];
 8         for(int i=0;i<p.length();i++){
 9             word[p.charAt(i)]++;
10         }
11         while(right<s.length()){
12             if(word[s.charAt(right++)]-->=1) count--;
13             if(count==0) res.add(left);
14             if(right-left==p.length()&&word[s.charAt(left++)]++>=0) count++;
15         }
16         return res;
17     }
18 }

 

posted @ 2017-03-19 01:32  CodesKiller  阅读(139)  评论(0编辑  收藏  举报