The task is taken from leetcode
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
My solution
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var findMedianSortedArrays = function(nums1, nums2) {
let i1 = i2 = 0;
let smallest, valBefore;
const pivot = Math.floor((nums1.length + nums2.length) / 2);
while (nums1[i1] || nums2[i2]) {
smallest = (nums2[i2] === void 0 || nums1[i1] < nums2[i2])
? nums1[i1++]
: nums2[i2++];
if ((nums1.length + nums2.length) % 2) {
if (pivot === i1 + i2 - 1) {
return smallest;
}
} else {
if (pivot - 1 === i1 + i2 - 1) {
valBefore = smallest
}
if (pivot === i1 + i2 - 1) {
return (smallest + valBefore) / 2;
}
}
}
};
This is the first idea that I had and it reached 99th percentile. I guess you can't take this evaluation seriously. And I'm sure there's a better solution.
[-1,0,1,2,3,4,5], thennums1[1]||nums2[1]will be false, and the loop will exit. \$\endgroup\$