438. 匹配字符串(顺序不同但个数相同的字符串) Find All Anagrams in a String

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".

  1. public class Solution {
  2. public List<int> FindAnagrams(string s, string p) {
  3. List<int> list = new List<int>();
  4. if (string.IsNullOrWhiteSpace(s) || s.Length < p.Length)
  5. return list;
  6. int sum = 0;
  7. int[] pArray = new int[26];
  8. foreach (char c in p){
  9. sum = sum + (c - 'a');
  10. ++pArray[c - 'a'];
  11. }
  12. int sum1 = 0;
  13. int startIndex = 0;
  14. for (int endIndex = 0; endIndex < s.Length; endIndex++){
  15. if (pArray[s[endIndex] - 'a'] == 0){
  16. sum1 = 0;
  17. startIndex = endIndex + 1;
  18. continue;
  19. }
  20. sum1 += s[endIndex] - 'a';
  21. if (endIndex - startIndex + 1 == p.Length){
  22. if(sum1 == sum) list.Add(startIndex);
  23. sum1 -= s[startIndex++] - 'a';
  24. }
  25. }
  26. return list;
  27. }
  28. }





原文地址:https://www.cnblogs.com/xiejunzhao/p/dc81ba095635e1f575ca737521c2ecc5.html