(C language) Rotate array

Given an array, rotate the elements of the array k positions to the right, where k is a non-negative number. Source: LeetCode

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
Rotate right 1 step: [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]

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

void reverse(int* nums,int i,int j){//颠倒函数
     int temp=0;//用来做中间值进行交换
      while(i<j){//定义两个下标一个指向第一位,一个指向最后一位,交换后,一个后移,
//另一个前移,直到两个相遇,完成全部颠倒
         //进行交换
         temp=nums[i];
         nums[i]=nums[j];
         nums[j]=temp;
         i++;//后移
         j--;//前移
     }
 }
 void rotate(int* nums, int numsSize, int k){//当我们要将一个数组元素向右移k位时只
//需要将前n-k个元素颠倒,再将后k个颠倒,再将全部元素颠倒,就可以完成转换了.
      if(k%numsSize==0 || numsSize==1){//判断数组中元素为1或0则直接返回.
         return;
     }
     reverse(nums,0,numsSize-(k%numsSize)-1);//颠倒前n-k个元素
     reverse(nums,numsSize-(k%numsSize),numsSize-1);//颠倒后k个元素
     reverse(nums,0,numsSize-1);//颠倒数组中所有元素完成轮转
 }

 int main() {
     int nums[] = { 1, 2, 3, 4, 5, 6, 7 };
     int k = 4;//右移次数
     rotate(nums, sizeof(nums) / sizeof(int), k);
     for (int i = 0; i < sizeof(nums) / sizeof(int); i++) {//打印轮转后的数组
         printf("%d ", nums[i]);
     }
     return 0;
 }