4. Median of Two Sorted Arrays

官方的鏈接：4. Median of Two Sorted Arrays

Description ：

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)).

Example1:

nums1 = [1, 3]

nums2 = [2]

The median is 2.0

Example2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

問題描述

給定長度分別為m和n的兩個排序數組nums1和nums2，在時間復雜度O(log (m+n))內算出數組的中位數

思路

兩個有序數組的中位數和Top K問題類似。這裏從小到大直接定位第k個，簡單理解為從nums1中獲取第i個，而從nums2中獲取第j=k-i個，其中i=0~k。當定位到nums[i-1]<=nums2[j]和nums[j-1]<=nums[i]即完成，當然還有邊界問題。

把中位數也當作是top k問題，最後進行奇偶判斷。詳情可參考這裏Share my O(log(min(m,n)) solution with explanation，代碼也是借鑒的。

值得註意的是幾個邊界判斷問題，比如i為0或者m，j為0或者n。

[github-here]

 1 public class Q4_MedianOfTwoSortedArrays {
 2     public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 3 
 4         int m = nums1.length;
 5         int n = nums2.length;
 6         // make sure that m <= n
 7         if (m > n) {
 8
 
             return findMedianSortedArrays(nums2, nums1);
 9         }
10         // n>=m,i = 0 ~ m,so j = (m + n + 1) / 2 -i > 0.
11         int i = 0, j = 0, imax = m, imin = 0, halfLen = (m + n + 1) / 2;
12         int maxLeft = 0, minRight = 0;
13         while (imin <= imax) {
14             i = (imin + imax) / 2;
15             j = halfLen - i;
16             if (i < m && nums2[j - 1] > nums1[i]) {
17                 imin = i + 1;
18             } else if (i > 0 && nums1[i - 1] > nums2[j]) {
19                 imax = i - 1;
20             } else {
21                 if (i == 0) {
22                     //the target is in nums2
23                     maxLeft = nums2[j - 1];
24                 } else if (j == 0) {
25                     //the target is in nums1
26                     maxLeft =  nums1[i - 1];
27                 } else {
28                     maxLeft = Math.max(nums1[i - 1], nums2[j - 1]);
29                 }
30                 break;
31             }
32         }
33         //odd
34         if ((m + n) % 2 == 1) {
35             return (double)maxLeft;
36         }
37         //even
38         if (i == m) {
39             //nums1 is out of index m
40             minRight = nums2[j];
41         } else if (j == n) {
42             //nums2 is out of index n
43             minRight = nums1[i];
44         } else {
45             //others
46             minRight = Math.min(nums1[i], nums2[j]);
47         }
48         return (double) (maxLeft + minRight) / 2;
49     }
50 
51     public static void main(String[] args) {
52         new Q4_MedianOfTwoSortedArrays().findMedianSortedArrays(new int[] { 1, 3 }, new int[] { 2 });
53 
54     }
55 }

Q4：Median of Two Sorted Arrays