## 题目

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

Example 1:

```Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]
```

Constraints:

• `1 <= nums.length <= 105`
• `-231 <= nums[i] <= 231 - 1`
• `0 <= k <= 105`

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

## 思路

`[1,2,3,4,5,6,7] =>  [5,6,7,1,2,3,4]`

```[1,2,3,4,5,6,7] =>  [7，6，5，4，3，2，1] 1）整体翻转

[7,6,5] => [5, 6, 7]
[4, 3, 2, 1] => [1, 2, 3, 4]```

## 代码

``````class Solution {
public void rotate(int[] nums, int k) {
int offset = k%nums.length;
if (offset == 0) return;
reverse(nums, 0, nums.length -1);
reverse(nums, 0, offset-1);
reverse(nums, offset, nums.length -1);
}

private void reverse(int[] nums, int left, int right) {
while(left < right) {
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
}
}``````
Published in数据结构和算法