4. Median of Two Sorted Arrays leetcode python New season for 2016
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)).
the main idea is if nums1_mid < nums2_mid nums1 = nums1_right_part nums2 = nums2_left_part
we need constant space and O(log(m+n)) time
class Solution(object):
def findMedianSortedArrays(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: float
"""
nums1_length = len(nums1)
nums2_length = len(nums2)
left = (nums1_length + nums2_length + 1) / 2
right = (nums1_length + nums2_length + 2) / 2
return (self.getKth(nums1, 0, nums2, 0, left) + self.getKth(nums1, 0, nums2, 0, right)) / 2.0
#if nums1 mid < nums2 mid keep nums1 right + nums2 left
def getKth(self, nums1, nums1_start, nums2, nums2_start, k):
if nums1_start > len(nums1) - 1:
return nums2[nums2_start + k - 1]
if nums2_start > len(nums2) - 1:
return nums1[nums1_start + k - 1]
if k == 1:
return min(nums1[nums1_start], nums2[nums2_start])
nums1_mid, nums2_mid = 1 << 31, 1 << 31
if nums1_start + k / 2 - 1 < len(nums1):
nums1_mid = nums1[nums1_start + k / 2 - 1]
if nums2_start + k / 2 - 1 < len(nums2):
nums2_mid = nums2[nums2_start + k / 2 - 1]
if nums1_mid < nums2_mid:
return self.getKth(nums1, nums1_start + k / 2, nums2, nums2_start, k - k / 2)
else:
return self.getKth(nums1, nums1_start, nums2, nums2_start + k / 2, k - k / 2)