1.5 Palindrome Permutation

Description

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

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

Method

count the frequency of each character, if there has more than one character's frequency is odd number, it will be false ,otherwise it is true;

Time and Space Complexity

Time : o(n) Space : o(n)

Code

public class Solution {

  public boolean canPermutePalindrome(String s) {
       HashMap<Character, Integer> map = new HashMap<>();
       int count  = 0;
       for (char c : s.toCharArray()){
            if (map.containsKey(c)){
                map.put(c, 1 + map.get(c));
            } else {
                map.put(c, 1);
            }
       }
       for (Integer i : map.values()){
            if (i % 2 != 0){
                count++;
            }
       }
       if (count > 1){
           return false;
       }
       return true;
}

}