Leetcode: 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.



[show hint]



Hint:

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

Naive想法就是保存一个原数组的拷贝，然后把原数组分成前len-k个元素和后k个元素两部分，把后k个元素放到前len-k个元素前面去。这样做需要O(N)空间

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可能比数组长度len要大，所以要先 k%=len；