LeetCode 4, Median of Two Sorted Arrays

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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)).（有两个有序数组nums1和nums2，他们的长度分别是m,n，找到两个数组的中位数，并且时间复杂度为O(log(m+n))

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {    int keyPosition=0;    int SUM=nums1Size+nums2Size;    double answer=0;    if(SUM%2==0)        keyPosition=SUM/2;    else keyPosition=(SUM+1)/2;    int i=0,j=0,count=0;    bool flag=true;    bool flag1=true;    char c;    if(nums1Size==0)    {            if(SUM%2==0)                {                    answer=(nums2[SUM/2]+nums2[SUM/2-1]);                    answer/=2;                }            else answer=nums2[SUM/2];            return answer;    }    else if(nums2Size==0)    {            if(SUM%2==0)                {                    answer=(nums1[SUM/2]+nums1[SUM/2-1]);                    answer/=2;                }            else answer=nums1[SUM/2];            return answer;    }   while(flag)    {        while(flag1&&i<nums1Size&&j<nums2Size&&nums1[i]<=nums2[j])        {            i++;            count++;            if(count==keyPosition)            {                c='i';                flag=false;                flag1=false;                break;            }        }        while(flag1&&j<nums2Size&&i<nums1Size&&nums2[j]<=nums1[i])        {            j++;            count++;            if(count==keyPosition)            {                c='j';                flag=false;                flag1=false;                break;            }        }        if(i==nums1Size)        {                if(SUM%2==0)                        {                            if(!flag)                                {                                    answer=nums1[i-1]+nums2[j];                                    answer/=2;                                }                                else                                 {                                     answer=(nums2[keyPosition-count-1+j]+nums2[keyPosition-count+j]);                                     answer/=2;                                }                        }                else                {                    if(!flag)                        answer=nums1[i-1];                    else answer=nums2[keyPosition-count-1+j];                }                return answer;        }        else if(j==nums2Size)        {               if(SUM%2==0)               {                   if(!flag)                   {                       answer=nums2[j-1]+nums1[i];                       answer/=2;                   }                   else                   {                        answer=(nums1[keyPosition-count-1+i]+nums1[keyPosition-count+i]);                        answer/=2;                   }               }                        else                 {                    if(!flag)                        answer=nums2[j-1];                    else answer=nums1[keyPosition-count-1+i];                }                return answer;        }    }    if(SUM%2==0)    {        if(c=='i')            {                answer=(nums1[i-1]+(nums1[i]<nums2[j]?nums1[i]:nums2[j]));                answer/=2;            }        else            {                answer=(nums2[j-1]+(nums2[j]<nums1[i]?nums2[j]:nums1[i]));                answer/=2;            }    }    else    {        if(c=='i')            answer=nums1[i-1];        else            answer=nums2[j-1];    }    return answer;}

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