Question

1470. 重新排列数组

English Version

题目描述

给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。

请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。

 

示例 1：

输入：nums = [2,5,1,3,4,7], n = 3
输出：[2,3,5,4,1,7] 
解释：由于 x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 ，所以答案为 [2,3,5,4,1,7]

示例 2：

输入：nums = [1,2,3,4,4,3,2,1], n = 4
输出：[1,4,2,3,3,2,4,1]

示例 3：

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

 

提示：

• 1 <= n <= 500
• nums.length == 2n
• 1 <= nums[i] <= 10^3

解法

方法一

class Solution:
  def shuffle(self, nums: List[int], n: int) -> List[int]:
    ans = []
    for i in range(n):
      ans.append(nums[i])
      ans.append(nums[i + n])
    return ans

class Solution {
  public int[] shuffle(int[] nums, int n) {
    int[] ans = new int[n << 1];
    for (int i = 0, j = 0; i < n; ++i) {
      ans[j++] = nums[i];
      ans[j++] = nums[i + n];
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> shuffle(vector<int>& nums, int n) {
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      ans.push_back(nums[i]);
      ans.push_back(nums[i + n]);
    }
    return ans;
  }
};

func shuffle(nums []int, n int) []int {
  var ans []int
  for i := 0; i < n; i++ {
    ans = append(ans, nums[i])
    ans = append(ans, nums[i+n])
  }
  return ans
}

function shuffle(nums: number[], n: number): number[] {
  let ans = [];
  for (let i = 0; i < n; i++) {
    ans.push(nums[i], nums[n + i]);
  }
  return ans;
}

impl Solution {
  pub fn shuffle(nums: Vec<i32>, n: i32) -> Vec<i32> {
    let n = n as usize;
    let mut res = Vec::new();
    for i in 0..n {
      res.push(nums[i]);
      res.push(nums[n + i]);
    }
    res
  }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* shuffle(int* nums, int numsSize, int n, int* returnSize) {
  int* res = (int*) malloc(sizeof(int) * n * 2);
  for (int i = 0; i < n; i++) {
    res[2 * i] = nums[i];
    res[2 * i + 1] = nums[i + n];
  }
  *returnSize = n * 2;
  return res;
}

方法二

class Solution:
  def shuffle(self, nums: List[int], n: int) -> List[int]:
    nums[::2], nums[1::2] = nums[:n], nums[n:]
    return nums

impl Solution {
  pub fn shuffle(mut nums: Vec<i32>, n: i32) -> Vec<i32> {
    let n = n as usize;
    for i in 0..n * 2 {
      let mut j = i;
      while nums[i] > 0 {
        j = if j < n { 2 * j } else { 2 * (j - n) + 1 };
        nums.swap(i, j);
        nums[j] *= -1;
      }
    }
    for i in 0..n * 2 {
      nums[i] *= -1;
    }
    nums
  }
}