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)).

Example 1:

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nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:

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nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

这个题目一看要求的时间复杂度为O(long(m+n))，基本说明这个题目要用二分法来做。

该题的做法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]，那么所求第k小一定不在 A[0] ~ A[k/2 - 1] 中。

代码来自九章算法: 

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public double findMedianSortedArrays(int A[], int B[]) {
    int len = A.length + B.length;
    if (len % 2 == 1) {
        return findKth(A, 0, B, 0, len / 2 + 1);
    }
    return (findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)) / 2.0;
}
// find kth number of two sorted array
public static int findKth(int[] A, int A_start,
                          int[] B, int B_start,
                          int k) {
    if (A_start >= A.length) {
        return B[B_start + k - 1];
    }
    if (B_start >= B.length) {
        return A[A_start + k - 1];
    }
    if (k == 1) {
        return Math.min(A[A_start], B[B_start]);
    }
    int A_key = A_start + k / 2 - 1 < A.length
            ? A[A_start + k / 2 - 1]
            : Integer.MAX_VALUE;
    int B_key = B_start + k / 2 - 1 < B.length
            ? B[B_start + k / 2 - 1]
            : Integer.MAX_VALUE;
    if (A_key < B_key) {
        return findKth(A, A_start + k / 2, B, B_start, k - k / 2);
    } else {
        return findKth(A, A_start, B, B_start + k / 2, k - k / 2);
    }
}