# 1470. Shuffle the Array

## Description

Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn].

Return the array in the form [x1,y1,x2,y2,...,xn,yn].

Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7]
Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].


Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]


Example 3:

Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]


Constraints:

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

## Solutions

• 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;
}
};

• 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


• 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
}
}