LeetCode-in-All
4. Median of Two Sorted Arrays
Hard
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
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
Solution
object Solution {
@SuppressWarnings(Array("scala:S3776"))
def findMedianSortedArrays(nums1: Array[Int], nums2: Array[Int]): Double = {
if (nums2.length < nums1.length) {
findMedianSortedArrays(nums2, nums1)
} else {
val n1 = nums1.length
val n2 = nums2.length
var cut1: Int = 0
var cut2: Int = 0
var low: Int = 0
var high: Int = n1
while (low <= high) {
cut1 = (low + high) / 2
cut2 = (n1 + n2 + 1) / 2 - cut1
val l1 = if (cut1 == 0) Int.MinValue else nums1(cut1 - 1)
val l2 = if (cut2 == 0) Int.MinValue else nums2(cut2 - 1)
val r1 = if (cut1 == n1) Int.MaxValue else nums1(cut1)
val r2 = if (cut2 == n2) Int.MaxValue else nums2(cut2)
if (l1 <= r2 && l2 <= r1) {
if ((n1 + n2) % 2 == 0) {
return (math.max(l1, l2) + math.min(r1, r2)) / 2.0
}
return math.max(l1, l2)
} else if (l1 > r2) {
high = cut1 - 1
} else {
low = cut1 + 1
}
}
0.0
}
}
}