266. Palindrome Permutation

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

For example,
"code" -> False, "aab" -> True, "carerac" -> True.

这道题收到了Palindrome Permutation2的启发，用了odd检验odd-even>1否。代码如下：

public class Solution {
    public boolean canPermutePalindrome(String s) {
        Map<Character,Integer> map = new HashMap<>();
        int odd = 0;
        for(char c:s.toCharArray()){
            map.put(c,map.getOrDefault(c,0)+1);
            odd+=map.get(c)%2==1?1:-1;
        }
        return odd>1?false:true;
    }
}

当然，本题也可以不用计数器来做，用hashSet来做，其实说白了，就是看出现奇数的char是否>1,代码如下：

public class Solution {
    public boolean canPermutePalindrome(String s) {
        Set<Character> set = new HashSet<>();
        for(char c:s.toCharArray()){
            if(set.contains(c)){
                set.remove(c);
            }else{
                set.add(c);
            }
        }
        return set.size()>1?false:true;
    }
}