Given an array, rotate the array to the right by k steps, where k is non-negative.
Test Cases
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]
Solution
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n = nums.size();
k %= n;
reverse(nums, 0, n-k-1);
reverse(nums, n-k, n-1);
reverse(nums, 0, n-1);
}
void reverse(vector<int> &nums, int start, int end) {
while(start < end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
}
};class Solution {
public void rotate(int[] nums, int k) {
int n = nums.length;
k %= n;
reverse(nums, 0, n-k-1);
reverse(nums, n-k, n-1);
reverse(nums, 0, n-1);
}
public void reverse(int[] nums, int start, int end) {
while(start < end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
}
}class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
n = len(nums)
k %= n
self.reverse(nums, 0, n-k-1)
self.reverse(nums, n-k, n-1)
self.reverse(nums, 0, n-1)
def reverse(self, nums, start, end) -> None:
while start < end:
nums[start], nums[end] = nums[end], nums[start]
start+=1
end-=1