[leetcode ][4] Median of Two Sorted Arrays

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本文介绍了一种在两个已排序数组中找到中位数的方法，通过二分查找实现，确保了整体运行时间复杂度为O(log(m+n))。提供了一个Python实现示例。

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)).
https://leetcode.com/problems/median-of-two-sorted-arrays/
思路：利用二分查找

class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        m = len(nums1); n = len(nums2)
        if (m+n)%2 == 1:
            return self.binarySearch(nums1, m, nums2, n, (m+n)/2+1)
        else:
             return float((self.binarySearch(nums1, m, nums2, n, (m + n) / 2) + self.binarySearch(nums1, m, nums2, n, (m + n) / 2 + 1))) / 2

    def binarySearch(self, nums1, m, nums2, n, k):
        if m<=0:
            return nums2[k-1]
        if n<=0:
            return nums1[k-1]
        if k<=1:
            return min(nums1[0], nums2[0])
        if nums1[m/2] > nums2[n/2]:
            if (m/2 + n/2 +1 ) >= k:
                return self.binarySearch(nums1, m/2, nums2, n, k)
            else:
                return self.binarySearch(nums1, m, nums2[n/2+1:n], n-n/2-1, k - n/2 -1)
        else:
            if(m/2 + n/2 +1 ) >= k:
                return self.binarySearch(nums1, m, nums2, n/2, k)
            else:
                return self.binarySearch(nums1[m/2+1:m], m-m/2-1, nums2, n, k-m/2-1)