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
public class Solution {
public double FindMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums2.Length < nums1.Length) {
return FindMedianSortedArrays(nums2, nums1);
}
int cut1;
int cut2;
int n1 = nums1.Length;
int n2 = nums2.Length;
int low = 0;
int high = n1;
while (low <= high) {
cut1 = (low + high) / 2;
cut2 = ((n1 + n2 + 1) / 2) - cut1;
int l1 = cut1 == 0 ? int.MinValue : nums1[cut1 - 1];
int l2 = cut2 == 0 ? int.MinValue : nums2[cut2 - 1];
int r1 = cut1 == n1 ? int.MaxValue : nums1[cut1];
int r2 = cut2 == n2 ? int.MaxValue : 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;
}
}
return 0.0;
}
}