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:

先统计p中字符的个数，vector<int>（128,0）.再从s的开头开始每次找p字符串长度个字符，来验证是否相同。如果不相同出现了直接break，如果一直都相同了，则将起始位置加入结果res中。

 1 class Solution {
 2 public:
 3     vector<int> findAnagrams(string s, string p) {
 4         vector<int> res;
 5         if(s.empty()) return res;
 6         vector<int> cnt(128, 0);
 7         int ns = s.size(); 
 8         int np = p.size();
 9         for(int i = 0; i < np; i++){
10             cnt[p[i] - ‘a‘]++;
11             //cnt[p[i]]++;
12         }
13         int i = 0;
14         while(i <= ns - np){
15             vector<int> tmp = cnt;
16             bool success = true;
17             for(int j = i; j < i + np; j++){
18                 tmp[s[j] - ‘a‘]--;
19                 //tmp[s[j]]--;
20                 if(tmp[s[j] - ‘a‘] < 0){
21                      success = false;
22                      break;
23                 }
24             }
25             
26             if(success) res.push_back(i);
27             i++;
28         }
29         return res;
30     }
31 };

View Code

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".

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".