[LeetCode][JavaScript]Median of Two Sorted Arrays

Median of Two Sorted Arrays

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)).

https://leetcode.com/problems/median-of-two-sorted-arrays/

如果是奇数长度，中位数就是中间的数；如果是偶数长度，中位数是中间两个数的平均数。

难点是题目要求复杂度是O(m+n)。

我的解法是先算出两个数组的总长度，然后循环从排序过的两个数组中取比较小的那个，统计长度，当长度等于总长度的一半时，根据奇数和偶数两种情况得到结果。

 1 /**
 2  * @param {number[]} nums1
 3  * @param {number[]} nums2
 4  * @return {number}
 5  */
 6 var findMedianSortedArrays = function(nums1, nums2) {
 7     var i = 0, j = 0, count = 0, middleLeft = -1;
 8     var len = nums1.length + nums2.length;
 9     while(i < nums1.length || j < nums2.length){
10         var curr = 0;
11         if(nums2[j] === undefined || nums1[i] <= nums2[j]){
12             curr = nums1[i];
13         }else if(nums1[i] === undefined || nums1[i] > nums2[j]){
14             curr = nums2[j];
15         }
16         count++;
17         if(len % 2 !== 0){
18             if(count === (len - 1) / 2 + 1){
19                 return curr;
20             }
21         }else{
22             if(count === len / 2){
23                 middleLeft = curr;
24             }else if(count === (len / 2 + 1) && middleLeft !== -1){
25                 return (curr + middleLeft) / 2;
26             }
27         }
28
29         if(nums2[j] === undefined || nums1[i] <= nums2[j]){
30             i++;
31         }else if(nums1[i] === undefined || nums1[i] > nums2[j]){
32             j++;
33         }
34     }
35     return 0;
36 };