4. Median of Two Sorted Arrays
Given two sorted arrays nums1
and nums2
of size m
and n
respectively, return the median of the two sorted arrays.
Follow up: 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
Rust Solution
struct Solution;
impl Solution {
fn find_median_sorted_arrays(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
let n1 = nums1.len();
let n2 = nums2.len();
if n1 < n2 {
return Self::find_median_sorted_arrays(nums2, nums1);
}
let mut lo = 0;
let mut hi = n2 * 2;
while lo <= hi {
let mid2 = (lo + hi) / 2;
let mid1 = n1 + n2 - mid2;
let l1 = if mid1 == 0 {
std::i32::MIN
} else {
nums1[(mid1 - 1) / 2]
};
let l2 = if mid2 == 0 {
std::i32::MIN
} else {
nums2[(mid2 - 1) / 2]
};
let r1 = if mid1 == n1 * 2 {
std::i32::MAX
} else {
nums1[mid1 / 2]
};
let r2 = if mid2 == n2 * 2 {
std::i32::MAX
} else {
nums2[mid2 / 2]
};
if l1 > r2 {
lo = mid2 + 1;
} else if l2 > r1 {
hi = mid2 - 1;
} else {
return (l1.max(l2) + r1.min(r2)) as f64 / 2.0;
}
}
panic!()
}
}
#[test]
fn test() {
use assert_approx_eq::assert_approx_eq;
let nums1 = vec![1, 3];
let nums2 = vec![2];
let res = 2.0;
assert_approx_eq!(Solution::find_median_sorted_arrays(nums1, nums2), res);
let nums1 = vec![1, 2];
let nums2 = vec![3, 4];
let res = 2.5;
assert_approx_eq!(Solution::find_median_sorted_arrays(nums1, nums2), res);
let nums1 = vec![1];
let nums2 = vec![2, 3];
let res = 2.0;
assert_approx_eq!(Solution::find_median_sorted_arrays(nums1, nums2), res);
}
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