> For the complete documentation index, see [llms.txt](https://hannahpun.gitbook.io/leetcode-note/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hannahpun.gitbook.io/leetcode-note/array/189-rotate-array.md).

# # 189 Rotate Array

[LeetCode](https://leetcode.com/problems/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?

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function(nums, k) {}
```

### 怎麼解

會用 % 餘數，然後用移掉後面的接回來前面

```
var rotate = function(nums, k) {
  k = k%nums.length;
  
  for(let i = 0; i < nums.length - k; i++) {
      nums.push(nums.shift())
  }
  return nums
  
};
```

better

```javascript

let len = nums.length;
let newArr = new Array(len-1)
    

 for(let i=0; i<len ; i++){
     const newIndex = i + k;
     if(newIndex < len){
        newArr[newIndex] = nums[i];
     } else {
        newArr[newIndex%len] = nums[i];
     }
        
}

for(let i=0; i<len ; i++){
    nums[i] = newArr[i]
}

```


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