Median of Two Sorted Arrays 解答

Question

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)).

Solution 1 -- Traverse Array

Use merge procedure of merge sort here. Keep track of count while comparing elements of two arrays. Note to consider odd / even situation.

Time complexity O(n), space cost O(1).

 1 public class Solution {
 2     public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 3         int length1 = nums1.length, length2 = nums2.length, total = length1 + length2;
 4         int index1 = total / 2, index2;
 5         // Consider two situations: (n + m) is odd, (n + m) is even
 6         if (total % 2 == 0)
 7             index2 = total / 2 - 1;
 8         else
 9             index2 = total / 2;
10         // Traverse once to get median
11         int p1 = 0, p2 = 0, p = -1, tmp, median1 = 0, median2 = 0;
12         while (p1 < length1 && p2 < length2) {
13             if (nums1[p1] < nums2[p2]) {
14                 tmp = nums1[p1];
15                 p1++;
16             } else {
17                 tmp = nums2[p2];
18                 p2++;
19             }
20             p++;
21             if (p == index1)
22                 median1 = tmp;
23             if (p == index2)
24                 median2 = tmp;
25         }
26         if (p < index1 || p < index2) {
27             while (p1 < length1) {
28                 tmp = nums1[p1];
29                 p1++;
30                 p++;
31                 if (p == index1)
32                     median1 = tmp;
33                 if (p == index2)
34                     median2 = tmp;
35             }
36             while (p2 < length2) {
37                 tmp = nums2[p2];
38                 p2++;
39                 p++;
40                 if (p == index1)
41                     median1 = tmp;
42                 if (p == index2)
43                     median2 = tmp;
44             }
45         }
46         return ((double)median1 + (double)median2) / 2;
47     }
48 }

Solution 2 -- Binary Search

General way to find Kth element in two sorted arrays. Time complexity O(log(n + m)).

 1 public class Solution {
 2     public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 3         int m = nums1.length, n = nums2.length;
 4         if ((m + n) %2 == 1)
 5             return findKthElement(nums1, nums2, (m + n) / 2, 0, m - 1, 0, n - 1);
 6         else
 7             return (findKthElement(nums1, nums2, (m + n) / 2, 0, m - 1, 0, n - 1) * 0.5 + 
 8                     findKthElement(nums1, nums2, (m + n) / 2 - 1, 0, m - 1, 0, n - 1) * 0.5);
 9     }
10     
11     private double findKthElement(int[] A, int[] B, int k, int startA, int endA, int startB, int endB) {
12         int l1 = endA - startA + 1;
13         int l2 = endB - startB + 1;
14         if (l1 == 0)
15             return B[k + startB];
16         if (l2 == 0)
17             return A[k + startA];
18         if (k == 0)
19             return A[startA] > B[startB] ? B[startB] : A[startA];
20         int midA = k * l1 / (l1 + l2);
21         // Note here
22         int midB = k - midA - 1;
23         midA = midA + startA;
24         midB = midB + startB;
25         if (A[midA] < B[midB]) {
26             k = k - (midA - startA + 1);
27             endB = midB;
28             startA = midA + 1;
29             return findKthElement(A, B, k, startA, endA, startB, endB);
30         } else if (A[midA] > B[midB]) {
31             k = k - (midB - startB + 1);
32             endA = midA;
33             startB = midB + 1;
34             return findKthElement(A, B, k, startA, endA, startB, endB);
35         } else {
36             return A[midA];
37         }
38     }
39 }