LeetCode #438 - Find All Anagrams in a String

题目描述：

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".

找出字符串的所有的子字符串，使得它们是目标字符串的异位字符串。直接以字符串中每个字符作为起点，然后判断，需要O(n*p.size())，但是利用滑动窗口可以降低到O(n)的时间复杂度。用left，right变量标记当前子字符串的左右位置，定义一个变量count表示子字符串和p之间相差的字母数，用哈希表存储子字符串和p中每个字母的次数差距。每次循环先移动right，并更新哈希表和count，同时保持子字符串的长度不能超过p，否则就要移动left减小子字符串的长度。当子字符串已经包含p的所有字母之后，此时count为0，就得到了一个结果。

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> result;
        if(s.size()==0||p.size()==0) return result;
        int left=0;
        int right=0;
        int count=p.size();
        vector<int> hash(256,0);
        for(int i=0;i<p.size();i++) hash[p[i]]++;
        
        while(right<s.size())
        {
            #判断s[right]是否为p中元素，并更新count和hash,然后移动right
            if(hash[s[right]]>=1) count--;
            hash[s[right]]--;
            right++;
            #当count为0，说明已经找到了一个结果
            if(count==0) result.push_back(left);
            
            #当滑动窗口的长度已经等于p的长度时，需要更新count和hash，然后移动left
            if(right-left==p.size())
            {
                if(hash[s[left]]>=0) count++;
                hash[s[left]]++;
                left++;
            }
        }
        return result;
    }
};