438. Find All Anagrams in a String

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：https://leetcode.com/problems/find-all-anagrams-in-a-string/#/description

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。

public class Solution {
public List<Integer> findAnagrams(String s, String p) {
    List<Integer> list = new ArrayList<>();
    if (s == null || s.length() == 0 || p == null || p.length() == 0) return list;
    int[] hash = new int[256]; //character hash
    //record each character in p to hash
    for (char c : p.toCharArray()) {
        hash[c]++;
    }
    //two points, initialize count to p's length
    int left = 0, right = 0, count = p.length();
    while (right < s.length()) {
        //move right everytime, if the character exists in p's hash, decrease the count
        //current hash value >= 1 means the character is existing in p
        if (hash[s.charAt(right)] >= 1) count--;
        hash[s.charAt(right)]--;
        right++;
        
        //when the count is down to 0, means we found the right anagram
        //then add window's left to result list
        if (count == 0) list.add(left);
    
        //if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window
        //++ to reset the hash because we kicked out the left
        //only increase the count if the character is in p
        //the count >= 0 indicate it was original in the hash, cuz it won't go below 0
        if (right - left == p.length()) {
            if (hash[s.charAt(left)] >= 0) count++;
            hash[s.charAt(left)]++;
            left++;
        }
    }
    return list;
}
}

别人的相关总结：

https://discuss.leetcode.com/topic/68976/sliding-window-algorithm-template-to-solve-all-the-leetcode-substring-search-problem

更多讨论（好像没有发现其他思路）

https://discuss.leetcode.com/category/563/find-all-anagrams-in-a-string