加入收藏 | 设为首页 | 会员中心 | 我要投稿 李大同 (https://www.lidatong.com.cn/)- 科技、建站、经验、云计算、5G、大数据,站长网!
当前位置: 首页 > 百科 > 正文

[Swift]LeetCode266.回文全排列 $ Palindrome Permutation

发布时间:2020-12-14 05:07:04 所属栏目:百科 来源:网络整理
导读:Given a string,determine if a permutation of the string could form a palindrome. For example, "code" ?- False,? "aab" ?- True,? "carerac" ?- True. Hint: Consider the palindromes of odd vs even length. What difference do you notice? Count t

Given a string,determine if a permutation of the string could form a palindrome.

For example,
"code"?-> False,?"aab"?-> True,?"carerac"?-> True.

Hint:

  1. Consider the palindromes of odd vs even length. What difference do you notice?
  2. Count the frequency of each character.
  3. If each character occurs even number of times,then it must be a palindrome. How about character which occurs odd number of times?

给定一个字符串,确定该字符串的排列是否可以形成回文。

例如,

“code”->false,“aab”->true,“carerac”->true。

提示:

  1. 考虑奇数和偶数的回文长度。你注意到了什么区别?
  2. 计算每个字符的频率。
  3. 如果每个字符出现偶数次,那么它必须是回文。奇数次出现的字符怎么样?

 1 class Solution {
 2     func canPermutePalindrome(_ s:String) -> Bool {
 3         var t:Set<Character> = Set<Character>()
 4         for a in s.characters
 5         {
 6             if !t.contains(a)
 7             {
 8                 t.insert(a)
 9             }
10             else
11             {
12                 t.remove(a)
13             }
14         }
15         return t.isEmpty || t.count == 1
16     }
17 }

(编辑:李大同)

【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容!

    推荐文章
      热点阅读