Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
Follow up: 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.
Example 3:
Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:
Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:
Input: nums1 = [2], nums2 = []
Output: 2.00000
Constraints:
nums1.length == mnums2.length == n0 <= m <= 10000 <= n <= 10001 <= m + n <= 2000-106 <= nums1[i], nums2[i] <= 106
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class Solution:
def findMedianSortedArrays(
self,
nums1: List[int],
nums2: List[int]
) -> float:
l = len(nums1) + len(nums2)
if l % 2 == 1:
return self.kth(nums1, nums2, l // 2)
return (
self.kth(nums1, nums2, l // 2)
+ self.kth(nums1, nums2, l // 2 - 1)
) / 2
def kth(
self,
a: List[int],
b: List[int],
k: int
) -> int:
if not a:
return b[k]
if not b:
return a[k]
ia, ib = len(a) // 2, len(b) //2
ma, mb = a[ia], b[ib]
if ia + ib < k:
# remove from the front
if ma > mb:
return self.kth(a, b[ib+1:], k - ib - 1)
return self.kth(a[ia+1:], b, k - ia - 1)
else:
# remove from the back
if ma > mb:
return self.kth(a[:ia], b, k)
return self.kth(a, b[:ib], k)
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func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
m := len(nums1)
n := len(nums2)
if m > n {
nums1, nums2 = nums2, nums1
m, n = n, m
}
iMin := 0
iMax := m
halfLen := (m + n + 1) / 2
for iMin <= iMax {
i := (iMin + iMax) / 2
j := halfLen - i
if i < iMax && nums2[j-1] > nums1[i] {
iMin = i + 1
} else if i > iMin && nums1[i-1] > nums2[j] {
iMax = i - 1
} else {
maxLeft := 0
if i == 0 {
maxLeft = nums2[j-1]
} else if j == 0 {
maxLeft = nums1[i-1]
} else {
maxLeft = Max(nums1[i-1], nums2[j-1])
}
if (m+n)%2 == 1 {
return float64(maxLeft)
}
minRight := 0
if i == m {
minRight = nums2[j]
} else if j == n {
minRight = nums1[i]
} else {
minRight = Min(nums2[j], nums1[i])
}
return float64(maxLeft+minRight) / 2.0
}
}
return 0.0
}
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