# Rotate Array

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**?

## Analysis

孪生兄弟题：[Rotate String](https://aaronice.gitbook.io/lintcode/string/rotate-string)

### 3-step reverse 三步翻转法

经典的应用场景，左移或者右移都适用。需要注意的是把位移量k先对长度n取模，`k = k % n;`

### extra auxiliary array 额外辅助数组

`arr[(i + k) % nums.length] = nums[i];`

## Solution

### 3-step reverse - O(n) time, O(1) space

```java
class Solution {
    public void rotate(int[] nums, int k) {
        int n = nums.length;
        k = k % n; 
        reverse(nums, 0, n - k - 1);
        reverse(nums, n - k, n - 1);
        reverse(nums, 0, n - 1);
    }
    private void reverse(int[] nums, int s, int t) {
        while (s < t) {
            int tmp = nums[s];
            nums[s] = nums[t];
            nums[t] = tmp;
            s++;
            t--;
        }
    }
}
```

### external aux array - O(n) time, O(n) space

```java
class Solution {
    public void rotate(int[] nums, int k) {
        int[] arr = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            arr[(i + k) % nums.length] = nums[i];
        }
        for (int i = 0; i < nums.length; i++) {
            nums[i] = arr[i];
        }
    }
}
```

## Reference

<https://leetcode.com/problems/rotate-array/solution/>