求n的元素的最大最小值

public static int[] maxMin(int a[]) {
        int[] res = new int[2];
        int len = a.length;
        if (len <= 0) {
            return res;
        }
        res[0] = res[1] = a[0];
        if (len % 2 == 0) {
            for (int i = 0; i < len - 1; i += 2) {
                if (a[i] > a[i + 1]) {
                    int tem = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = tem;
                }
            }
            for (int i = 0; i < len; i += 2) {
                if (res[0] > a[i]) {
                    res[0] = a[i];
                }
            }
            for (int i = 1; i < len; i += 2) {
                if (res[1] < a[i]) {
                    res[1] = a[i];
                }
            }
        } else {
            for (int i = 0; i < len - 1 - 1; i += 2) {
                if (a[i] > a[i + 1]) {
                    int tem = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = tem;
                }
            }
            for (int i = 0; i < len - 1; i += 2) {
                if (res[0] > a[i]) {
                    res[0] = a[i];
                }
            }
            for (int i = 1; i < len - 1; i += 2) {
                if (res[1] < a[i]) {
                    res[1] = a[i];
                }
            }
            if (res[0] > a[len - 1]) {
                res[0] = a[len - 1];
            } else if (res[1] < a[len - 1]) {
                res[1] = a[len - 1];
            }
        }
        return res;
    }

两两比较的方法，大的在右边，小的在左边，然后在分别找最大最小值，n是偶数奇数要注意。