[LeetCode] Median of Two Sorted Arrays

There are two sorted arrays A and B 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)).

代码的主体比较好写，三个地方卡了我一会。一个是其中有一个长度可能为0，这时候要单独处理；而是直接int相加除以2，结果答案不对；第三点也是最烦的一点，就是这个第K大和下标的转换，数来数去的搞不清楚哪里应该+1，-1什么的，很烦。

class Solution {
public:
double findMedianSortedArrays(int A[], int m, int B[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
assert(A&&B);
assert(m>0||n>0);
if (m==0||n==0)
{
int * C=m==0?B:A;
int t=m==0?n:m;
return singleArray(C,t);
}
if ( m < n )
return solve(A,m,B,n,0,m-1);
else
return solve(B,n,A,m,0,n-1);
}
double solve(int a[],int m,int b[],int n,int l,int r)
{
int half=(m+n)>>1;
if ( l > r )
return solve(b,n,a,m,max(0,half-m),min(half,n-1));

int mid = (l+r)>>1;
int k= half-(mid);

if ( get(a,m,mid) > get(b,n,k) )
return solve(a,m,b,n,l,mid-1);
else if ( get(a,m,mid) < get(b,n,k-1) )
return solve(a,m,b,n,mid+1,r);
else
{
if ( (m+n)%2 == 0)
{
return (double(a[mid])+max(get(a,m,mid-1),get(b,n,k-1)))/2;
}
else
{
return a[mid];
}
}
}
int get(int a[],int n,int k)
{
if ( k<0 )
return INT_MIN;
else if ( k>=n)
return INT_MAX;
else
return a[k];
}
double singleArray(int* a ,int n)
{
assert(a&&n>0);
int mid=n>>1;
if ( n%2==0)
{
return (double(a[mid])+a[mid-1])/2;
}
else
return a[mid];
}
};