4. Median of Two Sorted Arrays

Posted 2020-10-07 carsonwuu

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)).

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

 1 class Solution {
 2     public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 3         int n1 = nums1.length;
 4         int n2 = nums2.length;
 5         if ((n1 + n2) % 2 == 1) {
 6             return findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2 + 1);
 7         } else {
 8             return (findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2) + findNumK(nums1, n1, nums2, n2, (n1 + n2) / 2 + 1))
 9                     / 2;
10         }
11     }
12 
13     public double findNumK(int[] nums1, int n1, int[] nums2, int n2, int k) {
14         if (n1 > n2) {
15             return findNumK(nums2, n2, nums1, n1, k);
16         }
17         if (n1 == 0) {
18             return nums2[k - 1];
19         }
20         if (k == 1) {
21             return Math.min(nums1[0], nums2[0]);
22         }
23         int temp1 = Math.min(k / 2, n1);
24         int temp2 = k - temp1;
25         if (nums1[temp1 - 1] < nums2[temp2 - 1]) {
26             return findNumK(subArray(nums1, temp1), n1 - temp1, nums2, n2, k - temp1);
27         } else if (nums1[temp1 - 1] > nums2[temp2 - 1]) {
28             return findNumK(nums1, n1, subArray(nums2, temp2), n2 - temp2, k - temp2);
29         } else {
30             return nums1[temp1 - 1];
31         }
32     }
33 
34     public int[] subArray(int[] a, int b) {
35         int[] temp = new int[a.length - b];
36         for (int i = 0; i < a.length - b; i++) {
37             temp[i] = a[b + i];
38         }
39         return temp;
40     }
41 }