Median of Two Sorted Arrays

题目描述：

　　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)).

　　找两个有序数组的中位数，这并不难。可是题目把时间复杂度降到O(log(m+n))，难度就上去了。

我尝试解答，想到了把问题转换为求两个有序数组的第k小的数，奈何能力有限，最终铩羽而归。

下面贴出一个Accepted Code：

double findKth(int a[], int m, int b[], int n, int k)
{
//always assume that m is equal or smaller than n
if (m > n)
return findKth(b, n, a, m, k);
if (m == 0)
return b[k - 1];
if (k == 1)
return min(a[0], b[0]);
//divide k into two parts
int pa = min(k / 2, m), pb = k - pa;
if (a[pa - 1] < b[pb - 1])
return findKth(a + pa, m - pa, b, n, k - pa);
else if (a[pa - 1] > b[pb - 1])
return findKth(a, m, b + pb, n - pb, k - pb);
else
return a[pa - 1];
}

double findMedianSortedArrays(int A[], int m, int B[], int n)
{
int total = m + n;
if (total % 2 == 1)
return findKth(A, m, B, n, total / 2 + 1);
else
return (findKth(A, m, B, n, total / 2)
+ findKth(A, m, B, n, total / 2 + 1)) / 2;
}

如果想了解思路来源，可参考原文链接：/article/1536805.html
ps:

　　oj论坛上有人给出了复杂度更低的算法，但是思路没有上述方法清晰。