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
class Solution {
/**
* @param Integer[] $nums1
* @param Integer[] $nums2
* @return Float
*/
public function findMedianSortedArrays($nums1, $nums2) {
$nums3 = [];
$sum = count($nums1) + count($nums2);
$med = ($sum / 2);
if ($sum % 2 === 0) {
$index = [$med - 1, $med];
} else {
$index = [floor($med)];
}
$i = 0;
$i1 = 0;
$i2 = 0;
while ($i < $sum) {
if (!isset($nums1[$i1])) {
$nums3[] = $nums2[$i2];
$i2++;
$i++;
continue;
}
if (!isset($nums2[$i2])) {
$nums3[] = $nums1[$i1];
$i1++;
$i++;
continue;
}
if ($nums1[$i1] < $nums2[$i2]) {
$nums3[] = $nums1[$i1];
$i1++;
} else {
$nums3[] = $nums2[$i2];
$i2++;
}
$i++;
}
return isset($index[1])
? ($nums3[$index[0]] + $nums3[$index[1]]) / 2
: $nums3[$index[0]];
}
}