266. Palindrome Permutation

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

Example

Input: s = "code"
Output: False

Input: s = "carerac"
Output: True
Explanation: 
"carerac" --> "carerac"
*/

public class Solution {
    /**
     * @param s: the given string
     * @return: if a permutation of the string could form a palindrome
     */
    public boolean canPermutePalindrome(String s) {
        HashMap<Character,Integer> map = new HashMap<>();
        int length = s.length();
        for (int i=0;i<length;i++) {
            char c = s.charAt(i);
            if (!map.containsKey(c)){
                map.put(c,1);
            } else {
                int val = map.get(c);
                val+=1;
                map.put(c,val);
            }
        }
        //计算val，只有两种情况下s为回文串
        //1.没有出现次数为基数的字母
        //2.string长度为基数，并有一个出现次数为基数的字母
        int oddCount = 0;
        for (Map.Entry<Character,Integer> entry : map.entrySet()) {
            int val = entry.getValue();
            if (val%2==1){
                oddCount++;
            }
        }
        return (oddCount==0 || (length%2==1 && oddCount==1));
    }
}