【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?
Related problem: Reverse Words in a String II

题目大意是将数组循环右移k位，循环右移的含义就是将数组向右移动，其最后一个数移动到数组第一个位置，倒数第二个数移动到最后一个位置。。。，这样算是完成了循环右移一位。
我的思路主要有两个：一、每次将数组的每一位向右移动一位，总共移动k次，但是这种做法会涉及到双层循环，当数组的长度比较大以及k的值比较大时，这种算法的效率低下，导致oj出现超时的现象，代码如下：

public class Solution {    public void rotate(int[] nums, int k) {        int i,j,temp;        int len = nums.length;        if(len == 0 || len == 1 || k == len)            return;        if(k > len)            k = k-len;        for(i = 0;i<k;++i) {            temp = nums[len-1];            for(j = len-1;j>0;j--) {                nums[j] = nums[j-1];            }            nums[0] = temp;        }    }}

二、利用三次逆序的原理，先将下标为0~n-k-1之间的元素逆序，再将n-k~n-1之间的元素逆序，最后将0~n-1之间的元素逆序即可（n为数组长度）。例如数组a[] = {1,2,3,4,5,6,7},循环右移三位。第一次逆序后变为{4,3,2,1,5,6,7},第二次逆序后变为{4,3,2,1,7,6,5}，第三次逆序后变为{5,6,7,1,2,3,4}，正是所求的数组，代码如下：

public class Solution {    public void rotate(int[] nums, int k) {        int len = nums.length;        if(len == 1 || k == len)            return;        if(k > len)            k = k-len;        reverse(nums, 0, len-k-1);        reverse(nums, len-k, len-1);        reverse(nums, 0, len-1);    }    public void reverse(int[] nums,int start,int end) {        if(start >= end)            return;        int mid = (end + start) / 2;        int i,j,temp;        for(i = start,j = end;i <= mid;++i,--j) {            temp = nums[i];            nums[i] = nums[j];            nums[j] = temp;        }    }}