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?
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通过观察,如果s是个even string,则要为Palindrome就不能有odd frequency的char, 如果s是 odd 则可以有一个odd frequency,所以总的来说至多可以有一个odd frequency 的char。
利用一个int 统计到底有多少个odd frequency,最后判断这个数是不是小于等于1。
class Solution {
public:
bool canPermutePalindrome(string s) {
int odd = 0;
int mp[256] = {0};
for(char c : s) odd += ((++mp[c] % 2) == 1) ? 1 : -1;
return odd<=1;
}
};