4.寻找两个有序数组的中位数

要求时间复杂度在O(log(m+n))

思路：二分。中位数是把数组分成左右两边个数相等的两部分。因此如果数组A的索引为i，则数组b的索引就为(m+n+1)/2-i。而我们的目标是找到合适的i，使得B[j-1]<=A[i]以及A[i-1]<=B[j]

class Solution(object):
    def findMedianSortedArrays(self, A, B):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        m, n = len(A), len(B)
        if m > n:
            A, B, m, n = B, A, n, m
        if n == 0:
            return None
        imin, imax, half_len = 0, m, (m + n + 1) / 2
        while imin <= imax:
            i = (imin + imax) / 2
            j = half_len - i
            if i < m and B[j-1] > A[i]:
                imin = i + 1
            elif i > 0 and A[i-1] > B[j]:
                imax = i - 1
            else:
                if i == 0: max_of_left = B[j-1]
                elif j == 0: max_of_left = A[i-1]
                else: max_of_left = max(A[i-1], B[j-1])

                if (m + n) % 2 == 1:
                    return max_of_left

                if i == m: min_of_right = B[j]
                elif j == n: min_of_right = A[i]
                else: min_of_right = min(A[i], B[j])

                return (max_of_left + min_of_right) / 2.0