Day6 merge two ordered arrays

// Here are two buttons   Non decreasing order   Array of arranged integers  nums1  and  nums2, There are two other integers  m  and  n , respectively  nums1  and  nums2  The number of elements in .
//
// Would you please   Merge  nums2  To  nums1  in , Make the merged array press   Non decreasing order   array .
//
// Be careful ： Final , The merged array should not be returned by the function , It's stored in an array  nums1  in . In response to this situation ,nums1  The initial length of is  m + n, The top  m  Elements represent the elements that should be merged , after  n  Elements are  0 , It should be ignored .nums2  The length of is  n .
// You can design and implement a time complexity of  O(m + n)


#include<stdio.h>
#include <time.h>
#include <cstdlib>
#include <string.h>
#include<vector>
#include<string> 
#include<map>
#include<algorithm>

// The most direct conclusion is ： Directly store the two arrays together , Sorting can be 
int main()
{
    
	int nums1[6] = {
     4, 0, 0, 0, 0, 0 }, m =1, nums2[5] = {
     1, 2, 3, 5, 6 }, n = 5;

	int nums_index = 0;
	for (int i = m; i < m+n; i++)
	{
    
		nums1[i] = nums2[nums_index];
		nums_index++;
	}

	// array （ Insertion method , Sort from small to large ）
	for (int i = 1; i < m+n; i++)
	{
    
		int t = nums1[i];
		int j = i - 1;
		while (j >= 0 && nums1[j] > t)
		{
    
			nums1[j + 1] = nums1[j];
			nums1[j] = t;
			j--;
		}
	}
	for (int i = 0; i < m + n; i++)
	{
    
		printf("%d ", nums1[i]);
	}



	return 0;
}

for leecode

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){
    
    int nums_index = 0;
    if(m != 0 & n != 0)
    {
    
        for (int i = m; i < m+n; i++)
        {
    
            nums1[i] = nums2[nums_index];
            nums_index++;
        }

            // array （ Insertion method , Sort from small to large ）
        for (int i = 1; i < m+n; i++)
        {
    
            int t = nums1[i];
            int j = i - 1;
            while (j >= 0 && nums1[j] > t)
            {
    
                nums1[j + 1] = nums1[j];
                nums1[j] = t;
                j--;
            }
        }
    }
    else if(m == 0 & n != 0)
    {
    
        for (int i = 0; i < m+n; i++)
        {
    
            nums1[i] = nums2[i];
        }
    }

}