4 Median of Two Sorted Arrays

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 == m
  • nums2.length == n
  • 0 <= m <= 1000
  • 0 <= n <= 1000
  • 1 <= m + n <= 2000
  • -106 <= nums1[i], nums2[i] <= 106
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
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
}