LeetCode in Net

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 == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= 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;
    }
}

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