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".
这道题要求统计原字符串中，与模板字符串构成Anagrams的子串有多少，所谓Anagrams，也就是在寻找pattern的一个全排列，但是这里不用这么麻烦，只需要统计patter中各个字母的出现次数，之后逐步滑动，判断原字符串中与子串等长的字符串个字母出现的次数，patter和源字符串统计结果相等，那就说明该位置匹配
1 class Solution { 2 public: 3 vector<int> findAnagrams(string s, string p) { 4 vector<int> sv(26, 0), pv(26, 0), res; 5 if (s.size() < p.size()) 6 return res; 7 for (int i = 0; i < p.size(); i++) 8 { 9 pv[p[i]-'a']++; 10 sv[s[i]-'a']++; 11 } 12 if (sv == pv) 13 res.push_back(0); 14 for (int i = p.size(); i < s.size(); i++) 15 { 16 sv[s[i]-'a']++; 17 sv[s[i-p.size()]-'a']--; 18 if (pv == sv) 19 res.push_back(i - p.size() + 1); 20 } 21 return res; 22 } 23 };
当然也可以做一些优化思路：
1 class Solution { 2 public: 3 vector<int> findAnagrams(string s, string p) { 4 if (s.empty()) return {}; 5 vector<int> res, m(256, 0); 6 int left = 0, right = 0, cnt = p.size(), n = s.size(); 7 for (char c : p) ++m[c]; 8 while (right < n) { 9 if (m[s[right++]]-- >= 1) --cnt; 10 if (cnt == 0) res.push_back(left); 11 if (right - left == p.size() && m[s[left++]]++ >= 0) ++cnt; 12 } 13 return res; 14 } 15 };
这种方法只需要遍历一遍并且不需要比较两个向量的大小