Description:
Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
**Explanation:**
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
**Explanation:** 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:

  • Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
  • Could you do it in-place with O(1) extra space?

题意：给定一个一维数组和一个整数k，将这个一维数组向右循环移动k步，即最后一个元素移动到数组的首部，其他的向后移动；要求空间复杂度为O(1)；

解法一：最简单的方法就是循环k次，每次向右移动一步，从数组尾部开始赋给它前一个元素的值，最后第一个元素赋给它最后一个元素的值；这里使用一个变量存储最后一个元素的值；

Java
class Solution {
    public void rotate(int[] nums, int k) {
        while (k-- > 0) {
            int temp = nums[nums.length - 1];
            for (int i = nums.length - 1; i > 0; i--) {
                nums[i] = nums[i - 1];
            }
            nums[0] = temp;
        }
    }
}

解法二：假设我们要旋转的数组为nums=[x1,x2,x3,x4,x5,x6,x7]，旋转的步数为k=3；那么我们可以这样求解；

  1. 将数组nums所有元素逆序，得到nums=[x7,x6,x5,x4,x3,x2,x1]
  2. 将数组前k个元素逆序，得到nums= [x5,x6,x7,x4,x3,x2,x1]
  3. 将数组的后nums.length - k个元素逆序，得到nums=[x5,x6,x7,x1,x2,x3,x4]，即最终的答案
class Solution
    public void rotate(int[] nums, int k) {
        k %= nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
    }

    private void reverse(int[] nums, int st, int ed) {
        while