力扣练习1

1.数组串联问题

就将一个n长度的数组变成2n，并将里面的值再复制一份放进去。

Java:

class Solution { public int[] getConcatenation(int[] nums) { //创建新数组 int l=nums.length; int[]ans=new int [2*l]; for(int i=0;i<l;i++){ ans[i]=nums[i]; ans[i+l]=nums[i]; } return ans; } }

C

int* getConcatenation(int* nums, int numsSize, int* returnSize) { int n=numsSize; *returnSize=2*n; int *ans=(int*)malloc(sizeof(int)*2*n); for(int i=0;i<n;i++){ ans[i]=nums[i]; ans[i+n]=nums[i]; } return ans; }

C++

class Solution { public: vector<int> getConcatenation(vector<int>& nums) { vector<int> ans = nums; ans.insert(ans.end(), nums.begin(), nums.end()); return ans;} };

#include <vector> using namespace std; class Solution { public: vector<int> getConcatenation(vector<int>& nums) { int n = nums.size(); vector<int> ans(2 * n); for (int i = 0; i < n; i++) { ans[i] = nums[i]; ans[i + n] = nums[i]; } return ans; } };

python

class Solution(object): def getConcatenation(self, nums): return nums+nums def getConcatenation(nums): n = len(nums) ans = [0] * (2 * n) for i in range(n): ans[i] = nums[i] ans[i + n] = nums[i] return ans

python3

class Solution: def getConcatenation(self, nums: list[int]) -> list[int]: return nums + nums class Solution: def getConcatenation(self, nums: list[int]) -> list[int]: n = len(nums) # 创建长度 2n 的数组 ans = [0] * (2 * n) for i in range(n): ans[i] = nums[i] ans[i + n] = nums[i] return ans

2.重新排列数组

Java

class Solution { public int[] shuffle(int[] nums, int n) { int arr[]=new int[2*n]; int j=0; for(int i=0;i<n;i++){ arr[j++]=nums[i]; arr[j++]=nums[i+n]; } return arr; } }

Python3

class Solution: def shuffle(self, nums: list[int], n: int) -> list[int]: res = [] # 新建空数组 for i in range(n): res.append(nums[i]) # 加入 x1, x2... res.append(nums[i + n]) # 加入 y1, y2... return res

C

int* shuffle(int* nums, int numsSize, int n, int* returnSize) { *returnSize = 2 * n; // 返回数组长度 int* res = (int*)malloc(2 * n * sizeof(int)); // 手动申请内存 for (int i = 0; i < n; i++) { res[2 * i] = nums[i]; // 偶数位放 x res[2 * i + 1] = nums[i + n]; // 奇数位放 y } return res; }

C++

class Solution { public: vector<int> shuffle(vector<int>& nums, int n) { vector<int> res; for (int i = 0; i < n; i++) { res.push_back(nums[i]); res.push_back(nums[i + n]); } return res; } };

C#

public class Solution { public int[] Shuffle(int[] nums, int n) { int[] res = new int[2 * n]; for (int i = 0; i < n; i++) { res[2 * i] = nums[i]; res[2 * i + 1] = nums[i + n]; } return res; } }