Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]
Constraints:
1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105
Follow up:
Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
Could you do it in-place with O(1) extra space?
Solutions
Solution 1: Reverse three times
We can assume the length of the array is $n$ and calculate the actual number of steps needed by taking the module of $k$ and $n$, which is $k \bmod n$.
Next, let us reverse three times to get the final result:
Reverse the entire array.
Reverse the first $k$ elements.
Reverse the last $n - k$ elements.
For example, for the array $[1, 2, 3, 4, 5, 6, 7]$, $k = 3$, $n = 7$, $k \bmod n = 3$.
In the first reverse, reverse the entire array. We get $[7, 6, 5, 4, 3, 2, 1]$.
In the second reverse, reverse the first $k$ elements. We get $[5, 6, 7, 4, 3, 2, 1]$.
In the third reverse, reverse the last $n - k$ elements. We get $[5, 6, 7, 1, 2, 3, 4]$, which is the final result.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
/** Do not return anything, modify nums in-place instead. */functionrotate(nums:number[],k:number):void{constn:number=nums.length;k%=n;constreverse=(i:number,j:number):void=>{for(;i<j;++i,--j){constt:number=nums[i];nums[i]=nums[j];nums[j]=t;}};reverse(0,n-1);reverse(0,k-1);reverse(k,n-1);}