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.
Example1:
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”.
Example2:
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”.
解题思路
- 首先就要统计字符串p中字符出现的次数
- 然后从s的开头开始，每次找p字符串长度个字符，来验证字符个数是否相同
- 如果不相同出现了直接 break(比如说出现了字符个数小于0的情况)，
- 如果一直都相同了，则将起始位置加入结果 res 中
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
if(s.size() == 0){
return vector<int>();
}
vector<int> c(128, 0);
for(auto & i : p){
c[i]++;
}
int i = 0;
vector<int> res;
while(i < s.size()){
bool isFound = true;
vector<int> copyC(c);
for(int j = i; j < i + p.size(); j++){
if(--copyC[s[j]] < 0){
isFound = false;
break;
}
}
if(isFound){
res.push_back(i);
}
i++;
}
return res;
}
};
