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280. Wiggle Sort πŸ”’

Description

Given an integer array nums, reorder it such that nums[0] <= nums[1] >= nums[2] <= nums[3]....

You may assume the input array always has a valid answer.

 

Example 1:

Input: nums = [3,5,2,1,6,4]
Output: [3,5,1,6,2,4]
Explanation: [1,6,2,5,3,4] is also accepted.

Example 2:

Input: nums = [6,6,5,6,3,8]
Output: [6,6,5,6,3,8]

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 0 <= nums[i] <= 104
  • It is guaranteed that there will be an answer for the given input nums.

 

Follow up: Could you solve the problem in O(n) time complexity?

Solutions

Solution 1

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class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        for i in range(1, len(nums)):
            if (i % 2 == 1 and nums[i] < nums[i - 1]) or (
                i % 2 == 0 and nums[i] > nums[i - 1]
            ):
                nums[i], nums[i - 1] = nums[i - 1], nums[i]

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

    private void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

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

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func wiggleSort(nums []int) {
    for i := 1; i < len(nums); i++ {
        if (i%2 == 1 && nums[i] < nums[i-1]) || (i%2 == 0 && nums[i] > nums[i-1]) {
            nums[i], nums[i-1] = nums[i-1], nums[i]
        }
    }
}

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