【leetcode】438. Find All Anagrams in a String

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".

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思路：

运用hash表，用一个hash表h1统计字符串p中的元素，用另一个hash表h2统计字符串s中的元素，设p中的元素个数是m个，则h2只统计连续的m个元素。所以当坐标>=i时，每往后遍历一个元素，先前的元素也要减一个。

代码如下：

 class Solution {

 public:

     vector<int> findAnagrams(string s, string p) {

         vector<int>ans;

         int freq[]={};

         for(char c:p)

             freq[c-'a']++;

         for(int l=,r=,cnt=;r<s.size();++r)

         {

             if(r-l==p.length() && ++freq[s[l++]-'a']>)

                 cnt--;

             if(freq[s[r]-'a']--> && ++cnt==p.length())

                 ans.push_back(l);

         }

         return ans;

     }

 };

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