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).
PythonJavaC++GoTypeScriptRustJavaScriptC#
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)(code-box)
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;
}
}
}(code-box)
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());
}
};(code-box)
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)
}(code-box)
/**
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);
}(code-box)
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();
}
}(code-box)
/**
* @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);
};(code-box)
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;
}
}
}(code-box)
