Last Updated on 2023-02-06 by Clay
題目
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]
題目給定一組數值陣列的輸入,我們要把前半部認為是 x 陣列,後半認為是 y 陣列,接著把該數值陣列按照 [x1, y1, x2, y2... xn, yn]
的順序重新排列。
解題思路
Brute Force
最簡單的方法就是建立一個空陣列,接著設定一個 i
從 0 到 n-1 的迴圈,每次都依序把 nums[i]
跟 nums[n+i]
存入空陣列,就完成了題目要求的重新排列(洗牌)。
Brute Force 複雜度
只跑一次 for 迴圈、但是卻需要 n 個空間重新儲存。
Time Complexity | O(n) |
Space Complexity | O(n) |
C++ 範例程式碼
class Solution {
public:
vector<int> shuffle(vector<int>& nums, int n) {
// Init
vector<int> shuffleNums;
// Shuffle
for (int i=0; i<n; ++i) {
shuffleNums.emplace_back(nums[i]);
shuffleNums.emplace_back(nums[n+i]);
}
return shuffleNums;
}
};
Python 範例程式碼
class Solution:
def shuffle(self, nums: List[int], n: int) -> List[int]:
# Init
shuffle_nums = []
# Shuffle
for i in range(n):
shuffle_nums.append(nums[i])
shuffle_nums.append(nums[n+i])
return shuffle_nums