280. Wiggle Sort

    /*
     * 280. Wiggle Sort
     * 2016-6-27 by Mingyang
     * Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3]....
     * 对于题目的理解要准确！就是奇数位上的数字要大于等于周边的，那么我只用看小于的情况就交换好了
     */
    public void wiggleSort(int[] nums) {
        for (int i = 0; i < nums.length; i++)
            if (i % 2 == 1) {
                if (nums[i - 1] > nums[i])
                    swap(nums, i);
            } else if (i != 0 && nums[i - 1] < nums[i])
                swap(nums, i);
    }

    public void swap(int[] nums, int i) {
        int tmp = nums[i];
        nums[i] = nums[i - 1];
        nums[i - 1] = tmp;
    }

    // 下面是我拙劣的解法，不是in place的空间
    public static void wiggleSort11(int[] nums) {
        int len = nums.length;
        if (nums == null || len == 0)
            return;
        List<Integer> list1 = new ArrayList<Integer>();
        for (int i = 0; i < len; i++) {
            list1.add(nums[i]);
        }
        Collections.sort(list1);
        int start = 0;
        int end = list1.size() - 1;
        int count = 0;
        while (start < end) {
            nums[count++] = list1.get(start);
            nums[count++] = list1.get(end);
            start++;
            end--;
        }
        if (len % 2 != 0) {
            nums[count] = list1.get(start);
        }
    }