Rotate Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an array, 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
C++ Rotate Array LeetCode Solution
class Solution
{
public:
void rotate(int nums[], int n, int k)
{
if ((n == 0) || (k <= 0))
{
return;
}
// Make a copy of nums
vector<int> numsCopy(n);
for (int i = 0; i < n; i++)
{
numsCopy[i] = nums[i];
}
// Rotate the elements.
for (int i = 0; i < n; i++)
{
nums[(i + k)%n] = numsCopy[i];
}
}
};
Java Rotate Array LeetCode Solution
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 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--;
}
}
Python 3 Rotate Array LeetCode Solution
class Solution:
# @param nums, a list of integer
# @param k, num of steps
# @return nothing, please modify the nums list in-place.
def rotate(self, nums, k):
n = len(nums)
k = k % n
nums[:] = nums[n-k:] + nums[:n-k]
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