题目：

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".

链接：

3/25/2017

抄答案

注意的错误：

1. 如果用set和hashmap注意window长度内可能有重复的字符

2. pattern里会出现重复的字符

因为这两个错误，尤其是第二个，我的答案都错了，也想不出其他方法。

别人的

解决的问题：

1. count什么情况－：字符是需要的时候（本身在pattern里，而且还不到我们需要的数目）。count什么情况＋：最左边被left过滤掉的字符是我们需要的，也是下面right向右需要找的。

2. hash当中的值就是我们还缺的

3. 我们的window除了开始一直是定长，window当中很有可能有不需要的元素，需不需要担心？不需要，在这种情况下因为第一个判断条件不满足，count不会为0，不为0也就不会被加到结果中去。

4. 现在看着答案可以跟着思路理出来。但是如何将来还会想到呢？简单题吗？我觉得不是。

1 public class Solution { 2 public List
     
       findAnagrams(String s, String p) { 3     List
      
        list = new ArrayList<>(); 4     if (s == null || s.length() == 0 || p == null || p.length() == 0) return list; 5     int[] hash = new int[256]; //character hash 6     //record each character in p to hash 7     for (char c : p.toCharArray()) { 8         hash[c]++; 9     }10     //two points, initialize count to p's length11     int left = 0, right = 0, count = p.length();12     while (right < s.length()) {13         //move right everytime, if the character exists in p's hash, decrease the count14         //current hash value >= 1 means the character is existing in p15         if (hash[s.charAt(right)] >= 1) count--;16         hash[s.charAt(right)]--;17         right++;18         19         //when the count is down to 0, means we found the right anagram20         //then add window's left to result list21         if (count == 0) list.add(left);22     23         //if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window24         //++ to reset the hash because we kicked out the left25         //only increase the count if the character is in p26         //the count >= 0 indicate it was original in the hash, cuz it won't go below 027         if (right - left == p.length()) {28             if (hash[s.charAt(left)] >= 0) count++;29             hash[s.charAt(left)]++;30             left++;31         }32     }33     return list;34 }35 }
      
     

别人的相关总结：

更多讨论（好像没有发现其他思路）