The overall run time complexity should be O(log (m+n))
.
Example 1:
Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Constraints:
nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106
Table of Contents
Solution 1:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
nums = nums1 + nums2
nums = sorted(nums)
nums_len = len(nums)
if nums_len % 2 == 1:
return nums[int(nums_len/2)]
else:
return (nums[math.floor(nums_len/2)] + nums[math.floor(nums_len/2) - 1]) / 2
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
nums = nums1 + nums2
nums = sorted(nums)
nums_len = len(nums)
if nums_len % 2 == 1:
return nums[int(nums_len/2)]
else:
return (nums[math.floor(nums_len/2)] + nums[math.floor(nums_len/2) - 1]) / 2