# 189. 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?

#### 解题要点：

记录第一段和第二段，放在一个temp数组里，然后遍历把temp的数放到nums。

```python
class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        k = k % len(nums)
        firstHalf = []
        for i in range(len(nums)-k):
            firstHalf.append(nums[i])

        secondHalf = []
        for j in range(len(nums)-k, len(nums)):
            secondHalf.append(nums[j])

        temp = secondHalf + firstHalf

        for i in range(len(nums)):
            nums[i] = temp[i]
```

In-place && O(1) extra memory 解法：

```python
class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: None Do not return anything, modify nums in-place instead.
        """        
        
        def swap(nums, i, j):
            while i < j:
                temp = nums[i]
                nums[i] = nums[j]
                nums[j] = temp
                i += 1
                j -=1
                
        k = k % len(nums)
        swap(nums, 0, len(nums)-1)
        swap(nums, 0, k - 1)
        swap(nums, k, len(nums)-1)
```


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