[Leetcode]4. Median of Two Sorted Arrays @python

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)).

题目要求

有两个排好序的数组nums1，nums2，找到这两个数组的中位数，并要求时间复杂度为O(m+n)

解题思路

此题是借鉴了南郭子綦的解题思路。

将寻找中位数转化为找数组A，B中第k小的数。

代码

class Solution(object):

def getKth(self, A, B, k):
len_a,len_b = len(A),len(B)
# keep len(A) < len(B)
if len_a > len_b:
return self.getKth(B, A, k)
# when A is empty, return the kth minimal num in B
if len_a == 0:
return B[k - 1]
# if k == 1, return the minimal num between A[0] and B[0]
if k == 1:
return min(A[0],B[0])
pa = min(k / 2,len_a)
# we have make len(B) > len(A), so pb always less than or equal to len(B)
pb = k - pa
if A[pa - 1] <= B[pb - 1]:
return self.getKth(A[pa:],B,k - pa)
else:
return self.getKth(A,B[pb:],k - pb)

def findMedianSortedArrays(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: float
"""
if not nums1 and not nums2:
return None
n = len(nums1) + len(nums2)
if n % 2 == 1:
return self.getKth(nums1, nums2, n / 2 + 1)
else:
return (self.getKth(nums1, nums2, n / 2) + self.getKth(nums1, nums2, n / 2 + 1) ) * 0.5