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 lnums.length; int[]ansnew int [2*l]; for(int i0;il;i){ ans[i]nums[i]; ans[il]nums[i]; } return ans; } }Cint* getConcatenation(int* nums, int numsSize, int* returnSize) { int nnumsSize; *returnSize2*n; int *ans(int*)malloc(sizeof(int)*2*n); for(int i0;in;i){ ans[i]nums[i]; ans[in]nums[i]; } return ans; }Cclass Solution { public: vectorint getConcatenation(vectorint nums) { vectorint ans nums; ans.insert(ans.end(), nums.begin(), nums.end()); return ans;} };#include vector using namespace std; class Solution { public: vectorint getConcatenation(vectorint nums) { int n nums.size(); vectorint ans(2 * n); for (int i 0; i n; i) { ans[i] nums[i]; ans[i n] nums[i]; } return ans; } };pythonclass Solution(object): def getConcatenation(self, nums): return numsnums 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 anspython3class 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 ans2.重新排列数组Javaclass Solution { public int[] shuffle(int[] nums, int n) { int arr[]new int[2*n]; int j0; for(int i0;in;i){ arr[j]nums[i]; arr[j]nums[in]; } return arr; } }Python3class 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 resCint* 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; }Cclass Solution { public: vectorint shuffle(vectorint nums, int n) { vectorint 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; } }