algorithm@ O(3/2 n) time to findmaximum and minimum in a array

    public static int[] max_min(int[] a){
        //res[0] records the minimum value while res[1] records the maximal one.
        int res[] = new int[];
        int n = a.length;
        if(n == )  {
            return res;
        }
        if(n == ) {
            res[] = res[] = a[];
        }

        int min, max;
        if(n% == ) {
            res[] = res[] = a[];

            for(int i=; i<=n-; i+=) {
                if(a[i] < a[i+]) {
                    min = a[i];
                    max = a[i+];
                }
                else {
                    min = a[i+];
                    max = a[i];
                }
                res[] = Math.min(min, res[]);
                res[] = Math.max(max, res[]);
            }
        }
        else {
            res[] = Math.min(a[], a[]);
            res[] = Math.max(a[], a[]);
            for(int i=; i<=n-; i+=) {
                if(a[i] < a[i+]) {
                    min = a[i];
                    max = a[i+];
                }
                else {
                    min = a[i+];
                    max = a[i];
                }
                res[] = Math.min(min, res[]);
                res[] = Math.max(max, res[]);
            }
        }

        return res;
    }