[LeetCode266]Palindrome Permutation

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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通過觀察，如果s是個even string，則要爲Palindrome就不能有odd frequency的char， 如果s是 odd 則可以有一個odd frequency，所以總的來說至多可以有一個odd frequency 的char。

利用一個int 統計到底有多少個odd frequency，最後判斷這個數是不是小於等於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;
    }
};