problem:

There are two sorted arrays nums1 and nums2 of size m and n respectively.

有两个排序好的数列num1和num大小为m和n

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

找到这两个数列中的中位数,其时间复杂度为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

 1 class Solution(object):
 2     def findMedianSortedArrays(self, nums1, nums2):
 3         """
 4         :type nums1: List[int]
 5         :type nums2: List[int]
 6         :rtype: float
 7         """
 8
 9         m = len(nums1)
10         n = len(nums2)
11         if(m>n):
12             nums1,nums2,m,n = nums2,nums1,n,m
13         imin = 0
14         imax = m
15         halflen = (n+m+1)/2
16         while(imin <= imax):
17             i = (imin+imax)/2
18             j = halflen - i
19             if(i<imax and nums1[i]<nums2[j-1]):
20                 imin = i+1
21             elif(i>imin and nums2[j]<nums1[i-1]):
22                 imax = i-1
23             else:
24                 maxleft = 0
25                 if(i==0):
26                     maxleft = nums2[j-1]
27                 elif(j==0):
28                     maxleft = nums1[i-1]
29                 else:
30                     maxleft = max(nums1[i-1],nums2[j-1])
31                 if((n+m)%2==1):
32                     return maxleft
33                 minright = 0
34                 if(i == m):
35                     minright = nums2[j]
36                 elif(j == n):
37                     minright = nums1[i]
38                 else:
39                     minright = min(nums1[i],nums2[j])
40
41                 return (minright+maxleft)/2.0