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".
思路:
用一个滑动窗口,随时判断窗口里面的字符是否与p字符串相符合anagram。
vector<int> findAnagrams(string s, string p) { vector<int> pv(256,0), sv(256,0), res; if(s.size() < p.size()) return res; for(int i = 0; i < p.size(); ++i) { ++pv[p[i]]; ++sv[s[i]]; } if(pv == sv) res.push_back(0); for(int i = p.size(); i < s.size(); ++i) { ++sv[s[i]]; --sv[s[i-p.size()]]; if(pv == sv) res.push_back(i-p.size()+1); } return res; }
参考:
https://discuss.leetcode.com/topic/64390/c-o-n-sliding-window-concise-solution-with-explanation