C语言双指针多方法旋转数组解题LeetCode

void exchange(int* a, int* b)
{
int n=*a;
*a = *b;
*b = n;
}   //交换a，b位置
void reverse(int* nums,int left,int right)
{
while(left<right)
{
exchange(&nums[left],&nums[right]);
left++;
right--;
}   //对指定范围内元素进行翻转操作
}
void rotate(int* nums, int numsSize, int k){
k%=numsSize;
if(k==0)
{
return ;  //防止k过大或0导致无意义操作
}
reverse(nums,0,numsSize-1);//全倒序
reverse(nums,0,k-1);//前半部分倒序
reverse(nums,k,numsSize-1);//后半部分倒序
}

void rotate(int* nums, int numsSize, int k){
int arr[numsSize] = {0};
for(int n = 0;n<numsSize;n++)
{
arr[(k+n)%numsSize] = nums[n]; //nums所有元素向前移动 k 个单位，依次存到数组arr
}
for(int n = 0;n<numsSize;n++)
{
nums[n] = arr[n];  //将arr数组内容拷贝回原数组nums
}

  void * memcpy ( void * dest, const void * src, size_t num );

void rotate(int* nums, int numsSize, int k){
int arr[numsSize];
k %= numsSize;
memcpy(arr,nums+(numsSize-k),k*(int)sizeof(int));
memcpy(arr+k,nums,(numsSize-k)*(int)sizeof(int));
memcpy(nums,arr,numsSize*(int)sizeof(int));
}

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}
void swap(int* a, int* b) {
    int t = *a;
    *a = *b, *b = t;
}
void rotate(int* nums, int numsSize, int k) {
    k = k % numsSize;
    int count = gcd(k, numsSize);
    for (int start = 0; start < count; ++start) {
        int current = start;
        int prev = nums[start];
        do {
            int next = (current + k) % numsSize;
            swap(&nums[next], &prev);
            current = next;
        } while (start != current);
    }
}