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 == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -106 <= nums1[i], nums2[i] <= 106

Solution

#include <stdio.h>
#include <limits.h>

double findMedianSortedArrays(int* nums1, int n1, int* nums2, int n2) {
    if (n1 > n2) {
        return findMedianSortedArrays(nums2, n2, nums1, n1);
    }

    int low = 0, high = n1;
    while (low <= high) {
        int cut1 = (low + high) / 2;
        int cut2 = (n1 + n2 + 1) / 2 - cut1;

        int l1 = (cut1 == 0) ? INT_MIN : nums1[cut1 - 1];
        int l2 = (cut2 == 0) ? INT_MIN : nums2[cut2 - 1];
        int r1 = (cut1 == n1) ? INT_MAX : nums1[cut1];
        int r2 = (cut2 == n2) ? INT_MAX : nums2[cut2];

        if (l1 <= r2 && l2 <= r1) {
            if ((n1 + n2) % 2 == 0) {
                return (double)(fmax(l1, l2) + fmin(r1, r2)) / 2.0;
            } else {
                return (double)fmax(l1, l2);
            }
        } else if (l1 > r2) {
            high = cut1 - 1;
        } else {
            low = cut1 + 1;
        }
    }

    return 0.0;
}

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