LeeCode(4) Median of Two Sorted Arrays C語言版本

題目：
There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0
Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

思路：
由於數組是已經排序好的數組，則直接合並數組，然後根據兩數組總的元素個數之和，獲得中間數字。其中可以做一些優化。

C語言代碼：

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){

    double result =0;
    
    int sumSize = nums1Size + nums2Size;
    int len = sumSize / 2;
    int size =len+1;
   
    int * merge =(int *)malloc (size * sizeof(int)); 
    
    int i,n1,n2;
        
    for(i=0,n1 =0 ,n2 =0 ; i < size ; i++)
    {
        if(n1<nums1Size && n2<nums2Size)
        {
            if(nums1[n1] > nums2[n2])
            {
                merge[i] = nums2[n2];
                n2++;
            }else
            {
                merge[i] =nums1[n1];
                n1++;
            }
                        
        }
        
        else if(n1 >= nums1Size)
        {
            merge[i] = nums2[n2];
            n2++;
        }
        else if(n2 >= nums2Size)
        {
            merge[i] =nums1[n1];
            n1++;
        }
       
    }
    
  
    if(sumSize %2 !=0)
    {        
        result= merge[len];
    }
    else
    {
        result = merge[len-1] + merge[len];
        result = result /2;
    }
    
    return result;
}

效率：
Leecode中同一代碼執行的時間可能不同。本系列中，取代碼最佳效率。