两个有序数组的中位数

题目

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:

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

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

python

class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        if len(nums1) >= len(nums2):
            tem1 = nums1
            tem2 = nums2
        else:
            tem1 = nums2
            tem2 = nums1

        begin1 = 0
        end1 = len(tem1) - 1
        begin2 = 0
        end2 = len(tem2) - 1
        m = (end1 + end2 + 2) // 2
        l_length = 0
        n1=len(tem1)
        n2=len(tem2)
        while begin1 < end1 and begin2 < end2:
            middle = (begin2 + end2) // 2
            index = Solution.binary_search(tem1[begin1:end1 + 1], tem2[middle])
            l_length = (middle + 1 + index+begin1)
            if l_length > m:
                end1 = begin1 + index - 1
                end2 = middle
            elif l_length < m:
                begin1 += index
                begin2 = middle + 1
            else:
                x= tem1[begin1+index] if tem1[begin1+index]=0 and end2-begin2>=0:
            if end2-begin2 > end1-begin1:
                tem3 = tem2[begin2:end2 + 1]
                target = tem1[begin1]
            else:
                tem3 = tem1[begin1:end1 + 1]
                target = tem2[begin2]
            index = Solution.binary_search(tem3, target)
            l_length += index
            tem3.insert(index,target)
        else:
            tem3= tem2 if len(tem2)>len(tem1) else tem1
        if (n1+ n2) % 2 != 0:
            return tem3[position]
        else:
            x=tem3[position]+tem3[position-1]
            return x / 2.0

    @staticmethod
    def binary_search(num1, target):

        start = 0
        end = len(num1) - 1
        middle = (start + end) // 2
        while start < end:
            if num1[middle] > target:
                end = middle - 1
            elif num1[middle] < target:
                start = middle + 1
            else:
                return middle
            middle = (end + start) // 2
        if end < 0:
            return 0
        return end if num1[end] >= target else end + 1

思路二分查找的变形