# 189.Rotate-Array

## 189. Rotate Array

## 题目地址

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

## 题目描述

```
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?
```

## 代码

### Approach 1: Using Extra Array

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

      for (int i = 0; i < nums.length; i++) {
        nums[i] = a[i];
      }
    }
}
```

Approach #2 Using Cyclic Replacements **Confusion**

```java
class Solution {
  public void rotate(int[] nums, int k) {
    k = k % nums.length;
    int count = 0;
    for (int start = 0; count < nums.length; start++) {
      int current = start;
      int prev = nums[start];
      do {
        int next = (current + k) % nums.length;
        int temp = nums[next];
        nums[next] = prev;
        prev = temp;
        current = next;
        count++;
      } while(start != current);
    }
  }
}
```


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