rotate array

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note: Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

Hint:
Could you do it in-place with O(1) extra space?

Related problem: Reverse Words in a String II

#include<iostream>using namespace std;class Solution {public:void rotate(int nums[], int n, int k) /* solution one */{  //时间复杂度为O(k*n);空间复杂度O(1)int temp;for (int i = 0; i < k; i++){temp = nums[n - 1];for (int j = n - 2; j >= 0; j--){nums[j+1] = nums[j];}nums[0] = temp;}}void rotate2(int nums[], int n, int k) /* solution two */{k = k%n;int *temp = new int[n];memcpy(temp, nums + n - k, sizeof(int)*k);memcpy(temp+k,nums,sizeof(int)*(n-k));memcpy(nums, temp, sizeof(int)*n);delete[] temp;/*void *memcpy(void *dest, const void *src, size_t n);c和c++使用的内存拷贝函数，memcpy函数的功能是从源src所指的内存地址的起始位置开始拷贝n个字节到目标dest所指的内存地址的起始位置中。注意：dest和src在内存的位置，可能会产生覆盖的情况。*/}

      void reverse(int nums[], int start, int end)      {  int i = 0;  while (start < end)  {int tmp = nums[start];nums[start] = nums[end];nums[end] = tmp;start++;end--;  }      }     void rotateArray3(int nums[],int n, int k)     { //最优效率的方法，时间复杂度为O(n);空间复杂度为O(1)  reverse(nums, 0, n - k-1 );  reverse(nums, n - k, n-1);  reverse(nums, 0, n - 1);     }