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.[show hint]Hint:Could you do it in-place with O(1) extra space?

Naive想法就是保存一個原數組的拷貝，然后把原數組分成前len-k個元素和后k個元素兩部分，把后k個元素放到前len-k個元素前面去。這樣做需要O(N)空間

in-place做法是：

(1) reverse the array;

(2) reverse the first k elements;

(3) reverse the last n-k elements.

The first step moves the first n-k elements to the end, and moves the last k elements to the front. The next two steps put elements in the right order.

 1 public class Solution { 2     public void rotate(int[] nums, int k) { 3         int len = nums.length; 4         k %= len; 5         reverse(nums, 0, len-1); 6         reverse(nums, 0, k-1); 7         reverse(nums, k, len-1); 8     } 9     10     public void reverse(int[] nums, int l, int r) {11         while (l <= r) {12             int temp = nums[l];13             nums[l] = nums[r];14             nums[r] = temp;15             l++;16             r--;17         }18     }19 }

需要注意的是第4行，右移偏移量k可能比數組長度len要大，所以要先 k%=len；