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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 };
Permutation Palindrome
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