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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

Python3

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

Java

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;
    }
}

TypeScript

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;
}

C++

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;
    }
};

Go

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
}

C

/**
 * 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;
}

Rust

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
    }
}
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
    }
}

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