280. Wiggle Sort

Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3]....

For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4].

Solution:

The output array should meet these two conditions:

1. If index is odd, nums[index] >= nums[i- 1].

2. If index is even, nums[index] <= nums[i- 1].

Therefore, we go through the input array and at each index we check the above conditions.

If nums[i] does not meet the conditions, we swap it with the previous number.

By doing this, all the elements ahead of nums[i] still satisfy the conditions.

The time complexity is O(n).



Code:

public class Solution {
    public void wiggleSort(int[] nums) {
        for (int i = 1; i < nums.length; i++) {
            if ((i % 2 == 0 && nums[i] > nums[i - 1]) || 
                (i % 2 == 1 && nums[i] < nums[i - 1])) {
                
                int swap = nums[i];
                nums[i] = nums[i - 1];
                nums[i - 1] = swap;
            }
        }
    }
}