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)).
解题思路：
该题可以解决所有求有序数组A和B有序合并之后第k小的数！
该题的重要结论：
如果A[k/2-1]<B[k/2-1]，那么A[0]~A[k/2-1]一定在第k小的数的序列当中，可以用反证法证明。
具体的分析过程可以参考http://blog.csdn.net/zxzxy1988/article/details/8587244
class Solution {
public:
double findKth(int A[], int m, int B[], int n, int k)
{
//m is equal or smaller than n
if (m > n)
return findKth(B, n, A, m, k);
if (m == )
return B[k-];
if (k <= )
return min(A[], B[]);
int pa = min(k / , m), pb = k - pa;
if (A[pa-] < B[pb-])
{
return findKth(A + pa, m - pa, B, n, k - pa);
}
else if(A[pa-] > B[pb-])
{
return findKth(A, m, B + pb, n - pb, k - pb);
} else
return A[pa-];
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int k = m + n;
if (k & 0x1)
{
return findKth(A, m, B, n, k / + );
} else
{
return (findKth(A, m, B, n, k / ) + findKth(A, m, B, n, k / + )) / ;
}
}
};