Cracking the Coding Chpt 1.4 PalindromePermutation

ISsPalindromePermutation.java

// palindrome will have even letter counts and at most one odd letter count 
boolean isPalindromPermutation(String input) {
  	Map<String, Integer> frequency = new HashMap<String, Integer>(); 
  	for(String s : input.split("")) {
  		if(!frequency.containsKey(s)) {
  			frequency.put(s, 1); 
  		} else {
  			frequency.put(s, 1 + frequency.get(s)); 
  		}
  	}
  	
  	// at most one odd count 
  	int oddCount = 0; 
  	for(Map.Entry<String, Integer> entry : frequency.entrySet()) {
  		if(entry.getValue() % 2 != 0) {
  			++oddCount; 
  			if(oddCount > 1) {
  				return false; 
  			}
  		}
  	}
  	return true; 
  }