Description#
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 - 10 <= 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 $O(1)$.
1
2
3
4
5
6
7
8
9
10
11
12
| class Solution:
def rotate(self, nums: List[int], k: int) -> None:
def reverse(i: int, j: int):
while i < j:
nums[i], nums[j] = nums[j], nums[i]
i, j = i + 1, j - 1
n = len(nums)
k %= n
reverse(0, n - 1)
reverse(0, k - 1)
reverse(k, n - 1)
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| class Solution {
private int[] nums;
public void rotate(int[] nums, int k) {
this.nums = nums;
int n = nums.length;
k %= n;
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
}
private void reverse(int i, int j) {
for (; i < j; ++i, --j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
}
|
1
2
3
4
5
6
7
8
9
10
| class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n = nums.size();
k %= n;
reverse(nums.begin(), nums.end());
reverse(nums.begin(), nums.begin() + k);
reverse(nums.begin() + k, nums.end());
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
| func rotate(nums []int, k int) {
n := len(nums)
k %= n
reverse := func(i, j int) {
for ; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
}
reverse(0, n-1)
reverse(0, k-1)
reverse(k, n-1)
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| /**
Do not return anything, modify nums in-place instead.
*/
function rotate(nums: number[], k: number): void {
const n: number = nums.length;
k %= n;
const reverse = (i: number, j: number): void => {
for (; i < j; ++i, --j) {
const t: number = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
};
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
}
|
1
2
3
4
5
6
7
8
9
| impl Solution {
pub fn rotate(nums: &mut Vec<i32>, k: i32) {
let n = nums.len();
let k = (k as usize) % n;
nums.reverse();
nums[..k].reverse();
nums[k..].reverse();
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| /**
* @param {number[]} nums
* @param {number} k
* @return {void} Do not return anything, modify nums in-place instead.
*/
var rotate = function (nums, k) {
const n = nums.length;
k %= n;
const reverse = (i, j) => {
for (; i < j; ++i, --j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
};
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| public class Solution {
private int[] nums;
public void Rotate(int[] nums, int k) {
this.nums = nums;
int n = nums.Length;
k %= n;
reverse(0, n - 1);
reverse(0, k - 1);
reverse(k, n - 1);
}
private void reverse(int i, int j) {
for (; i < j; ++i, --j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
}
|
Solution 2#
1
2
3
4
| class Solution:
def rotate(self, nums: List[int], k: int) -> None:
k %= len(nums)
nums[:] = nums[-k:] + nums[:-k]
|
