Question

280. Wiggle Sort

中文文档

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

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]

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;
  }
}

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]);
      }
    }
  }
};

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]
    }
  }
}