438. Find All Anagrams in a String

1. Question

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

2. Examples

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".

3. Constraints

  • 1 <= s.length, p.length <= 3 * 104
  • s and p consist of lowercase English letters.

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/find-all-anagrams-in-a-string 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

利用数组进行计数,和给定的字符串计数的数组一致则保存索引。

class Solution {
  public List<Integer> findAnagrams(String s, String p) {
    List<Integer> ans = new ArrayList<>();

    if (s.length() < p.length()) {
      return ans;
    }

    int[] ps = new int[26];
    int[] res = new int[26];
    for (int i = 0; i < p.length(); i++) {
      ps[p.charAt(i) - 'a']++;
    }

    int left = 0;
    int right = p.length();

    for (int i = 0; i < p.length(); i++) {
      res[s.charAt(i) - 'a']++;
    }

    if (Arrays.equals(ps, res)) {
      ans.add(left);
    }

    while (right < s.length()) {
      res[s.charAt(right) - 'a']++;
      right++;

      res[s.charAt(left) - 'a']--;
      left++;

      if (Arrays.equals(ps, res)) {
        ans.add(left);
      }
    }

    return ans;
  }
}
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