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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
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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; } }
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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; } };
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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
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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 }
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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; }
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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 } }