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
#include <vector>
#include <algorithm>
#include <climits>
class Solution {
public:
double findMedianSortedArrays(std::vector<int>& nums1, std::vector<int>& nums2) {
if (nums2.size() < nums1.size()) {
return findMedianSortedArrays(nums2, nums1);
}
int n1 = nums1.size();
int n2 = nums2.size();
int low = 0;
int 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 (std::max(l1, l2) + std::min(r1, r2)) / 2.0;
}
return std::max(l1, l2);
} else if (l1 > r2) {
high = cut1 - 1;
} else {
low = cut1 + 1;
}
}
return 0.0;
}
};