438. Find All Anagrams in a String

原题

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".

代码实现

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;        }