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?

題意：一個數組向右“移動”k位

不額外開空間的情況下

這題有個巧妙的演算法，假設k小於length，那麼移動後 就是後k位移到前k位 (假設k為3吧)

1 2 3 4 5 6 7             

5 6 7 1 2 3 4(應該得到的結果)

我們把陣列分開，後面的k位和前面的

1 2 3 4   5 6 7

然後各部分翻轉

4 3 2 1   7 6 5

最後一起反轉，就能把後k位翻到前面去

5 6 7 1 2 3 4

class Solution {
    
 
public void rotate(int[] nums, int k) {
        if (nums.length == 0 || (k%(nums.length) == 0))
            return;
        int pos = k%nums.length;

        reverse(nums, 0, nums.length - pos - 1);
        reverse(nums, nums.length - pos, nums.length - 1);
        reverse(nums, 0, nums.length - 1);
    }
    public void reverse(int[] nums, int l, int r) {
        while (l < r) {
            int temp = nums[l];
            nums[l] = nums[r];
            nums[r] = temp;
            l++;
            r--;
        }
    }
}