LeetCode Rotate Array

Rotate Array Total Accepted: 5488 Total Submissions: 29811 My Submissions Question Solution 
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.

题目意思就是将数组按照指定的位置进行旋转，其中的Note说明至少有三种方法。

方法一：采用一个辅助空间，然后进行数据的拷贝，在此使用memcpy和memset等（面试有一些关于memcpy和memmove的题目，可以自行google），空间复杂度O(n)

void rotate(int nums[], int n, int k) {if (n <= 0 || k == 0)return;k > n ? k = k%n : n;const int intSize = sizeof(int);int * arr = new int[n];memset(arr, 0, n*intSize);memcpy(arr, nums, n*intSize);memcpy(nums, arr + (n-k), k*intSize);memcpy(nums + k, arr, (n - k)*intSize);delete[] arr;}

方法二：每个数字移动k次，空间复杂度O(1)，时间复杂度O(K*N)

for (int i = 0; i < k; i++)//每个数字移动k次{int tmp = nums[n - 1];for (int j = n - 1; j >= 0; j--)//后移数组元素{nums[j] = nums[j - 1];}nums[0] = tmp;}

方法三：根据规律旋转，时间复杂度和系统函数有关，空间复杂度O(0)

void rotate(int nums[], int n, int k) {reverse(nums, nums + n);//翻转整个数组[7,6,5,4,3,2,1]reverse(nums, nums + k%n);//翻转从起始位置到k%n位置处[5,6,7,4,3,2,1]reverse(nums + k%n, nums + n);//翻转剩余部分[5,6,7,1,2,3,4]}