README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1470.Shuffle%20the%20Array/README_EN.md	1120	Weekly Contest 192 Q1	Array

1470. Shuffle the Array

中文文档

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:

<li><code>1 &lt;= n &lt;= 500</code></li>

<li><code>nums.length == 2n</code></li>

<li><code>1 &lt;= nums[i] &lt;= 10^3</code></li>

Solutions

Solution 1: Simulation

We traverse the indices $i$ in the range $[0, n)$. Each time, we take $\textit{nums}[i]$ and $\textit{nums}[i+n]$ and place them sequentially into the answer array.

After the traversal is complete, we return the answer array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def shuffle(self, nums: List[int], n: int) -> List[int]:
        return [x for pair in zip(nums[:n], nums[n:]) for x in pair]

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;
    }
}

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

TypeScript

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

Rust

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

C

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