Count # of odd but not exact # of appearance Java


  • 0
    A

    Maybe this solution is too simple to be posted by you guys. Anyway, for someone who want to achieve this idea in Java, here it is. I use boolean[] instead of counting # of each character.

    public int longestPalindrome(String s) {
            int n = s.length();
            boolean[] odd = new boolean[60];
            for(char c : s.toCharArray())
                odd[c - 'A'] = !odd[c - 'A'];
            int odds = 0;
            for(boolean b : odd){
                if(b == true)
                    odds++;
            }
            return odds == 0 ? n : (n+1-odds);
        }
    

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