Leetcode-median of Sorted Arrays

Median of Sorted Arrays

The problem is to find median, the median number. This means that if the array size is even, it returns the average of the two numbers in the middle, if it is an odd number, which is the middle number.
The algorithm time efficiency requirements are O (log (m + N)), the specific ideas on the Internet are the same.

In addition, now Leetcode C + + arrays have been replaced by vectors, so we have to tidy up the use of vectors.
Reference: http://imatlab.lofter.com/post/286ffc_a81276
Http://www.cnblogs.com/wang7/archive/2012/04/27/2474138.html

C code

#include <iostream>using namespace STD;DoubleFindkth (intA[],intMintB[],intNintK) {if(M > N)returnFindkth (B, N, A, M, K);if(M = =0)returnb[k-1];if(k <=1)returna[0]<b[0]? a[0]:b[0];intia = (k/2< m)? k/2: M, ib = K-ia;if(a[ia-1] < b[ib-1]){returnFindkth (A+ia, M-ia, B, N, K-ia); }Else if(a[ia-1] > b[ib-1]){returnFindkth (A, M, B+ib, N-ib, K-ib); }Else{returna[ia-1]; }}DoubleFindmediansortedarrays (int* Nums1,intNums1size,int* NUMS2,intNums2size) {intK = nums1size + nums2size;if(K &0x1){returnFindkth (Nums1, Nums1size, Nums2, Nums2size, k/2+1); }Else{return(Findkth (Nums1, Nums2size, Nums2, Nums2size, k/2) + findkth (nums1, Nums1size, Nums2, Nums2size, k/2+1)/2); }}intMainintargcChar*argv[]) {intA[] = {1,3,5,7,9};intB[] = {2,4,6,8};Doubleresult = Findmediansortedarrays (A,5B4);printf("%d", (int) result);}

C + + code

Include <iostream>#include <vector>using namespace STD;classSolution { Public:DoubleFindmediansortedarrays ( vector<int>& Nums1, vector<int>& Nums2) {intm = Nums1.size ();intn = nums2.size ();intTotal = m + N;if(Total &0x1)returnFind_kth (Nums1, NUMS2, total/2+1);Else{return(Find_kth (Nums1, NUMS2, total/2) + find_kth (nums1, NUMS2, total/2+1))/2.0; }        }Private:intFind_kth ( vector<int>& A, vector<int>& B,intK) {intm = A.size ();intn = b.size ();if(M > N)returnFind_kth (B, A, K);if(M = =0)returnb[k-1];if(k = =1)returnMin (a[0], b[0]);//min (A, B) and Max (A, b) are available            intia = min (k/2, m), IB = K-ia;if(a[ia-1] < b[ib-1]) {a.erase (A.begin (), A.begin () +ia);returnFind_kth (A, B, K-ia); }Else if(a[ia-1] > b[ib-1]) {b.erase (B.begin (), B.begin () +ib);returnFind_kth (A, B, K-ib); }Else{returna[ia-1]; }        }};intMainintargcChar*argv[]) {intA[] = {1,3,5,7,9};intB[] = {2,4,6,8}; vector<int>A (&a[0], &a[5]);//Copy 5 elements     vector<int>B (&b[0], &b[4]);//Copy 4 elementsSolution solution;Doubleresult = Solution.findmediansortedarrays (A, B);printf("%d", (int) result);}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Leetcode-median of Sorted Arrays