# 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

### 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

``````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

``````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/

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