1.4 Palindrome Permutation

The basic idea of this problem is counting each character’s appearance time.

For even length string: each char should appear even times.

For odd length string: only one char can appear odd time.

I use bool oddCharAppeared to indicate if there was a char appear odd time.

void countFrequency(string& s , int* frequency){
int idx = -1;
for (const char & c : s)
{
idx = tolower(c) -'a';
if ( idx != -1 ) ++frequency[idx];
}
}

bool palindromePermutation(string& s){
int fre[ 26 ] = { 0 };
countFrequency(s,fre);
bool oddCharAppeared = false;
for (int i = 0; i<26; ++i) {
if(fre[i]%2 && !oddCharAppeared) oddCharAppeared = true;
else if (fre[i]%2 && oddCharAppeared) return false;
}
return true;
}