LeeCode_回文排列

class Solution {
    public boolean canPermutePalindrome(String s) {
        Map<Character, Integer> chars = new HashMap<>();
        for (char c : s.toCharArray()){
            if (chars.containsKey(c)){  // 判断该key在map中是否有key存在
                chars.put(c, chars.get(c)+1);
            }else{
                chars.put(c, 1);
            }
        }

        // 提出map遍历字符统计次数
        int Count = 0;
        for (int x :chars.values()){
            if (x%2 != 0){
                Count++;
            }
        }

        if (Count>1){      // 统计Count即字符个数为奇数的个数
            return false;
        }
        return true;
    }
}