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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；
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原文地址：http://www.cnblogs.com/EdwardLiu/p/4306556.html
