# -*- coding: utf8 -*-
'''
https://oj.leetcode.com/problems/median-of-two-sorted-arrays/
There are two sorted arrays A and B 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)).
===Comments by Dabay===
先计算中位数在合并后数组中的什么坐标为medium_pos。
同时考虑到,合并后数组的元素总数是奇数偶数的不同情况,用一个变量last来记录寻找到中位数的游标停下来的位置。
如果是奇数,这个last没有用;如果是偶数,用这个last和下一个可能的数求一个平均数。
'''
class Solution:
# @return a float
def findMedianSortedArrays(self, A, B):
len_a, len_b = len(A), len(B)
i = j = 0
medium_pos = (len_a + len_b) / 2
counter = 0
last = None
while counter < medium_pos:
if i < len_a and j < len_b:
if A[i] < B[j]:
last = A[i]
i = i + 1
else:
last = B[j]
j = j + 1
elif i < len_a:
last = A[i]
i = i + 1
elif j < len_b:
last = B[j]
j = j + 1
counter = counter + 1
if (len_a + len_b) % 2 == 0:
if i < len_a and j < len_b:
return (last + min(A[i], B[j])) / 2.0
elif i == len_a:
return (last + B[j]) / 2.0
elif j == len_b:
return (last + A[i]) / 2.0
else:
if i < len_a and j < len_b:
return min(A[i], B[j])
elif i == len_a:
return B[j]
elif j == len_b:
return A[i]
def main():
s = Solution()
A = [1, 2, 3, 4, 5]
B = []
print s.findMedianSortedArrays(A, B)
if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)