Question

324. 摆动排序 II

English Version

题目描述

给你一个整数数组 nums，将它重新排列成 nums[0] < nums[1] > nums[2] < nums[3]... 的顺序。

你可以假设所有输入数组都可以得到满足题目要求的结果。

 

示例 1：

输入：nums = [1,5,1,1,6,4]
输出：[1,6,1,5,1,4]
解释：[1,4,1,5,1,6] 同样是符合题目要求的结果，可以被判题程序接受。

示例 2：

输入：nums = [1,3,2,2,3,1]
输出：[2,3,1,3,1,2]

 

提示：

• 1 <= nums.length <= 5 * 104
• 0 <= nums[i] <= 5000
• 题目数据保证，对于给定的输入 nums ，总能产生满足题目要求的结果

 

进阶：你能用 O(n) 时间复杂度和 / 或原地 O(1) 额外空间来实现吗？

解法

方法一：排序

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

class Solution {
  public void wiggleSort(int[] nums) {
    int[] arr = nums.clone();
    Arrays.sort(arr);
    int n = nums.length;
    int i = (n - 1) >> 1, j = n - 1;
    for (int k = 0; k < n; ++k) {
      if (k % 2 == 0) {
        nums[k] = arr[i--];
      } else {
        nums[k] = arr[j--];
      }
    }
  }
}

class Solution {
public:
  void wiggleSort(vector<int>& nums) {
    vector<int> arr = nums;
    sort(arr.begin(), arr.end());
    int n = nums.size();
    int i = (n - 1) >> 1, j = n - 1;
    for (int k = 0; k < n; ++k) {
      if (k % 2 == 0)
        nums[k] = arr[i--];
      else
        nums[k] = arr[j--];
    }
  }
};

func wiggleSort(nums []int) {
  n := len(nums)
  arr := make([]int, n)
  copy(arr, nums)
  sort.Ints(arr)
  i, j := (n-1)>>1, n-1
  for k := 0; k < n; k++ {
    if k%2 == 0 {
      nums[k] = arr[i]
      i--
    } else {
      nums[k] = arr[j]
      j--
    }
  }
}

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var wiggleSort = function (nums) {
  let bucket = new Array(5001).fill(0);
  for (const v of nums) {
    bucket[v]++;
  }
  const n = nums.length;
  let j = 5000;
  for (let i = 1; i < n; i += 2) {
    while (bucket[j] == 0) {
      --j;
    }
    nums[i] = j;
    --bucket[j];
  }
  for (let i = 0; i < n; i += 2) {
    while (bucket[j] == 0) {
      --j;
    }
    nums[i] = j;
    --bucket[j];
  }
};

方法二：桶排序

class Solution:
  def wiggleSort(self, nums: List[int]) -> None:
    """
    Do not return anything, modify nums in-place instead.
    """
    bucket = [0] * 5001
    for v in nums:
      bucket[v] += 1
    n = len(nums)
    j = 5000
    for i in range(1, n, 2):
      while bucket[j] == 0:
        j -= 1
      nums[i] = j
      bucket[j] -= 1
    for i in range(0, n, 2):
      while bucket[j] == 0:
        j -= 1
      nums[i] = j
      bucket[j] -= 1

class Solution {
  public void wiggleSort(int[] nums) {
    int[] bucket = new int[5001];
    for (int v : nums) {
      ++bucket[v];
    }
    int n = nums.length;
    int j = 5000;
    for (int i = 1; i < n; i += 2) {
      while (bucket[j] == 0) {
        --j;
      }
      nums[i] = j;
      --bucket[j];
    }
    for (int i = 0; i < n; i += 2) {
      while (bucket[j] == 0) {
        --j;
      }
      nums[i] = j;
      --bucket[j];
    }
  }
}

class Solution {
public:
  void wiggleSort(vector<int>& nums) {
    vector<int> bucket(5001);
    for (int& v : nums) ++bucket[v];
    int n = nums.size();
    int j = 5000;
    for (int i = 1; i < n; i += 2) {
      while (bucket[j] == 0) --j;
      nums[i] = j;
      --bucket[j];
    }
    for (int i = 0; i < n; i += 2) {
      while (bucket[j] == 0) --j;
      nums[i] = j;
      --bucket[j];
    }
  }
};

func wiggleSort(nums []int) {
  bucket := make([]int, 5001)
  for _, v := range nums {
    bucket[v]++
  }
  n, j := len(nums), 5000
  for i := 1; i < n; i += 2 {
    for bucket[j] == 0 {
      j--
    }
    nums[i] = j
    bucket[j]--
  }
  for i := 0; i < n; i += 2 {
    for bucket[j] == 0 {
      j--
    }
    nums[i] = j
    bucket[j]--
  }
}