leetcode median of two sorted arrays

double findMedianSortedArrays2(int A[], int m, int B[], int n) {
if (m > n) return findMedianSortedArrays(B, n, A, m);
int minidx = 0, maxidx = m, i, j, num1, mid = (m + n + 1) >> 1, num2;
while (minidx <= maxidx)//实际上是找到A和B一起的前mid个数，i+j=mid，从而A[0]到[i-1]和B[0]到[j-1]一共就是mid个数
{
i = (minidx + maxidx) >> 1;
j = mid - i;
if (i<m && j>0 && B[j - 1] > A[i]) minidx = i + 1;
else if (i>0 && j<n && B[j] < A[i - 1]) maxidx = i - 1;
else
{
if (i == 0) num1 = B[j - 1];
else if (j == 0) num1 = A[i - 1];
else num1 = max(A[i - 1], B[j - 1]);
break;
}
}
if (((m + n) & 1)) return num1;
if (i == m) num2 = B[j];
else if (j == n) num2 = A[i];
else num2 = min(A[i], B[j]);
return (num1 + num2) / 2.;
}

int getkth(int s[], int m, int l[], int n, int k){
// let m <= n
if (m > n)
return getkth(l, n, s, m, k);
if (m == 0)
return l[k - 1];
if (k == 1)
return min(s[0], l[0]);

int i = min(m, k / 2), j = min(n, k / 2);
if (s[i - 1] > l[j - 1])
return getkth(s, m, l + j, n - j, k - j);
else
return getkth(s + i, m - i, l, n, k - i);
return 0;
}