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

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

这种题目很多方法，写了以下三种：

• 用一个set，放入char，重复（放不进去）拿出来，最后确认size，0或者1则是true，否则false
• 用一个boolean和hashset去跟踪
• 用一个array of int，放入，重复则index –，最后判断index的数字

public class Solution {
    public boolean canPermutePalindrome(String s) {
        Set<Character> set=new HashSet<Character>();
        for(int i=0; i<s.length(); ++i){
            if (!set.contains(s.charAt(i)))
                set.add(s.charAt(i));
            else 
                set.remove(s.charAt(i));
        }
        return set.size()==0 || set.size()==1;
    }
}

public class Solution {
    public boolean canPermutePalindrome(String s) {
        HashMap<Character,Integer> map = new HashMap<>();
        boolean res = false;
        for(int i=0;i<s.length();i++){
            if(map.get(s.charAt(i)) == null){
                map.put(s.charAt(i), 1);
                res = !res;
            }
            else{
                if(map.get(s.charAt(i)) == 0){
                    map.put(s.charAt(i), 1);
                    res = !res;
                }
                else{
                    map.put(s.charAt(i), 0);
                    res = !res;
                }
            }
            System.out.println("s:"+i+"res:"+res);
        }
        return res;
    }
}

 

public class Solution {
    public boolean canPermutePalindrome(String s) {
        int[] check = new int[128];
        int index = s.length();
        for(int i=0;i<s.length();i++){
            if(check[s.charAt(i) - 'a'] == 1){
                check[s.charAt(i) - 'a'] =0;
                index = index -2;
            }
            else{
                check[s.charAt(i) - 'a'] = 1;
            }
        }
        if(index ==1 || index == 0){
            return true;
        }
        return false;
    }
}