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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 the frequency of each character.
- 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:
- 考虑奇数和偶数的回文长度。你注意到了什么区别?
- 计算每个字符的频率。
- 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 }
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