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 <= n <= 500</code></li>
<li><code>nums.length == 2n</code></li>
<li><code>1 <= nums[i] <= 10^3</code></li>
Solutions
Solution 1: Simulation
We traverse the indices i in the range [0, n). Each time, we take nums[i] and 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 nums.
PythonJavaC++GoTypeScriptRustC
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](code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
/**
* 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;
}(code-box)
