Topic description:
Given an array, rotate the elements in the array K positions to the right, where K is a non-negative number, and return the new array
Example:
Input: [1,2,3,4,5,6,7],k=3;
Output: [5,6,7,1,2,3,4]
Method 1: Using an additional array, apply for an array of the same length as the original array to put each element in the correct position.
First put the last k values of the original array into the first k positions of the new array, and then put the remaining elements of the original array at the back.
void rotate(int* nums, int numsSize, int k)
{
int* new_nums = (int*)malloc(sizeof(int) * numsSize);
for (int i = 0; i < numsSize; i++)
{
new_nums[(i + k) % numsSize] = nums[i];
}
for (int i = 0; i < numsSize; i++)
{
nums[i] = new_nums[i];
}
free(new_nums);
Method 2: Array flip
Since we move the elements of the array to the right k times, the tail k elements will move to the head of the array, and the remaining elements will be k positions backward. Therefore, we can first flip all the elements so that the tail k elements are moved to the head of the array, and then we flip the first k elements and the rest to get the final answer.
Example:
Original array: 1 2 3 4 5 6 7
Flip all elements: 7 6 5 4 3 2 1
Flip the first k elements: 5 6 7 4 3 2 1
Then flip the remaining elements: 5 6 7 1 2 3 4
void Swap(int* a, int* b)
{
int tmp = *a;
*a = *b;
*b = tmp;
}
void Reverse(int* nums, int start, int end)
{
while (start < end)
{
Swap(&nums[start], &nums[end]);
start += 1;
end -= 1;
}
}
void rotate2(int* nums, int numsSize, int k)
{
k %= numsSize;
Reverse(nums, 0, numsSize - 1);
Reverse(nums, 0, k - 1);
Reverse(nums, k, numsSize - 1);
}
