[leetcode]4. Median of Two Sorted Arrays俩有序数组的中位数

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

Solution1: time complexity is O(m+n). merge two sorted array, then find the median

code:

 1 /*
 2   Time Complexity:  O(m+n)
 3   Space Complexity: O(m+n) 
 4 */
 5 
 6 class Solution {
 7     public double findMedianSortedArrays(int[] nums1, int[] nums2) {   // find median
 8         int[] mergedArray = mergeTwoSortedArray(nums1, nums2);
 9         int n = mergedArray.length; 
10         if( n % 2 == 0){
11             return (mergedArray[(n-1)/2] + mergedArray[n/2])/2.0; 
12         }else{
13             return mergedArray[n/2]; 
14         }
15     } 
16     
17     public int[] mergeTwoSortedArray(int[] nums1, int[] nums2){ // merge sort
18         int[] merged = new int[nums1.length + nums2.length];
19         int i = 0, j = 0, k = 0;
20         // 两个array同时扫
21         while( i < nums1.length && j < nums2.length){
22             merged[k++] = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++]; 
23         }
24         //扫到只剩下较长的那个array， either nums1 or nums2
25         while ( i < nums1.length ){
26              merged[k++] = nums1[i++];
27         }
28         while(j < nums2.length){
29             merged[k++] = nums2[j++];
30         }
31         return merged;
32     }
33 }

Solution2: time complexity is O(log (m+n)).

这个题的思路可以是找出2个sorted array 所有元素中，Kth 元素

因此我们可以把问题转化成，

“Given 2 sorted arrays, A, B of sizes m, n respectively, find the numbers which are NOT medians of the two sorted arrays”

如果我们能找到那些NOT medians的数字，删掉，并不断缩小范围，那最终剩下的一定就是actual median

Step1, 确定要找的median是merged之后array中的第几个元素

Step2, 用binary search切开nums1, nums2。发现nums1左半边的长度为4， nums2左半边的长度为2。

Step3, nums1左半边长度+ nums2左半边长度为6

Step4，所以第7个元素可能在nums1的右半边pivot上，或者在nums2的右半边pivot上

比较发现，第7个元素应该是7

code：

 1 /*
 2  Time Complexity: O(log(m+n))
 3  Space Complexity: O(log(m+n))
 4 */
 5 
 6 class Solution {
 7     public double findMedianSortedArrays(final int[] A, final int[] B) {
 8         int total = A.length + B.length;
 9         if (total %2 == 0){
10             return (findKth(A, 0, B, 0, total / 2) + findKth(A, 0, B, 0, total / 2 + 1)) / 2.0;
11         }else{
12             return findKth(A, 0, B, 0, total / 2 + 1);
13         }
14     }
15 
16     private static int findKth(final int[] A, int ai, final int[] B, int bi, int k) {
17         //always assume that A is shorter than B
18         if (A.length - ai > B.length - bi) {
19             return findKth(B, bi, A, ai, k);
20         }
21         if (A.length - ai == 0) return B[bi + k - 1];
22         if (k == 1) return Math.min(A[ai], B[bi]);
23 
24         //divide k into two parts
25         int k1 = Math.min(k / 2, A.length - ai), k2 = k - k1;
26         if (A[ai + k1 - 1] < B[bi + k2 - 1])
27             return findKth(A, ai + k1, B, bi, k - k1);
28         else if (A[ai + k1 - 1] > B[bi + k2 - 1])
29             return findKth(A, ai, B, bi + k2, k - k2);
30         else
31             return A[ai + k1 - 1];
32     }
33 }