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：

 public class Solution {

     public List<Integer> findAnagrams(String s, String p) {

         List<Integer> res = new ArrayList<Integer>();

         if(s==null||s.length()==0||p==null||p.length()==0) return res;

         int left = 0,right = 0;

         int count  =p.length();

         int[] word = new int[256];

         for(int i=0;i<p.length();i++){

             word[p.charAt(i)]++;

         }

         while(right<s.length()){

             if(word[s.charAt(right++)]-->=1) count--;

             if(count==0) res.add(left);

             if(right-left==p.length()&&word[s.charAt(left++)]++>=0) count++;

         }

         return res;

     }

 }

巴特西

438. Find All Anagrams in a Strin

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