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".这个算法的时间复杂度为O(n)，空间复杂度为O(1)。这个算法比较难理解，以后慢慢回头琢磨吧。
public IList<int> FindAnagrams(string s, string p){ IList<int> rtn = new List<int>(); int[] hash = new int[123]; //a~z: 97~122 foreach (var c in p){ hash[Convert.ToInt32(c)]++; //hash: key:char, value: occuring times } int eachBeg = 0, eachEnd = 0, count = p.Length; while (eachEnd < s.Length){ char tmpchar = s[eachEnd]; if (hash[tmpchar] >= 1) count--; hash[tmpchar]--; eachEnd++; //every time the eachEnd pointer to move toward right if (count == 0) rtn.Add(eachBeg); //reset the hash. if (eachEnd - eachBeg == p.Length){ char tmp = s[eachBeg]; if (hash[tmp] >= 0) count++; hash[tmp]++; eachBeg++; } } return rtn; }