Permutation Palindrome

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

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

Analyse: use hash map

 1 class Solution {
 2 public:
 3     bool canPermutePalindrome(string s) {
 4         unordered_map<char, int> um;
 5         for (char ch : s) 
 6             um[ch]++;
 7         
 8         bool canPermute = true;
 9         for (unordered_map<char, int>::iterator ite = um.begin(); ite != um.end(); ite++) {
10             if (ite->second % 2) {
11                 if (!canPermute) return false;
12                 else canPermute = false;
13             }
14         }
15         return true;
16     }
17 };

Analyse: use A[ch]. There are 128 characters in ASCII, we can convert a char to int automatically.

 1 class Solution {
 2 public:
 3     bool canPermutePalindrome(string s) {
 4         vector<bool> charOccurenceBool(128, false);
 5         for (char ch : s)
 6             charOccurenceBool[ch] = !charOccurenceBool[ch];
 7         
 8         int count = 0;
 9         for (int i = 0; i < charOccurenceBool.size(); i++) {
10             if (charOccurenceBool[i]) count++;
11         }
12         return count < 2;
13     }
14 };