Find All Anagrams in a String解题报告

Description:

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:

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".

Link:

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

解题方法：

用字符的分布来匹配anagrams，用slide window来优化暴力算法。当i>p的长度时，每次i++都相当于窗口右移动。左边需要减去刚刚划过的记录。

Tips:

不能用unordered_map来当哈希表。

Time Complexity：

O(n)

完整代码：

vector findAnagrams(string s, string p) 
    {
        vector result;
        int lenp = p.size();
        int lens = s.size();
        if(s.size() == 0 || p.size() > s.size())
            return result;
        vector hash1(26);
        vector hash2(26);
        for(auto a: p)
            hash1[a-'a']++;
        for(int i = 0; i < s.size(); i++)
        {
            hash2[s[i]-'a']++;
            if(i >= lenp) hash2[s[i-lenp]-'a']--;  //加上右边的记录，减去左边的记录
            if(hash1 == hash2) result.push_back(i-lenp+1);  
        }
        return result;   
    }