# 189. Rotate Array (1)

**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?

**Solution**

記住一個事實，shift k 之後的nums結果會等於下列三步驟所得的結果:

1. 先全部reverse
2. &#x20;reverse前面k個element {0...k-1}
3. &#x20;reverse後面剩下的element{k...n-1}

```
    public void rotate(int[] nums, int k) {
        //base on the fact:
        //result will be 
        // 1. the reverse of all nums
        // 2. reverse the the {0...k-1}
        // 3. reverse the remain {k...n-1}
        k = k % nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
    }
    
    public void reverse(int[] nums, int from, int end){
        int temp = 0;
        while(from < end){
            temp = nums[from];
            nums[from] = nums[end];
            nums[end] = temp;
            from++;
            end--;
        }
    }
```
