log(m+n)时间找到中位数

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昨天晚上我的室友出了一道题目：

现在有两个已经有序的数组A和B，A有m个元素，B有n个元素。现在使用log(m+n)的复杂度计算出它们的中位数。我大致想了一下，用分治算法。然后今天写出了代码：

/*This file implements : For two ordered arrays A with m integers and B with n integers.please find the median number within log(m + n) time.*/#include <stdio.h>void find_median_number(int A[], int start_1, int B[], int start_2, int kth_smallest, int size_of_A, int size_of_B) {int temp;int end_1;int end_2;int compare_from_A;int compare_from_B;int bound = 0;while (1) {bound = 0;if (start_1 >= size_of_A) {start_1 = size_of_A - 1;}if (start_2 >= size_of_B)  {start_2 = size_of_B - 1;}if (1 == kth_smallest) {temp = A[start_1] > B[start_2] ? B[start_2] : A[start_1];fprintf(stdout, " median : %d \n", temp);return;}end_1 = start_1 + (kth_smallest + 1) / 2 - 1;end_2 = start_2 + (kth_smallest + 1) / 2 - 1;if (end_1 >= size_of_A) {end_1 = size_of_A - 1;bound = 1;} if (end_2 >= size_of_B) {end_2 = size_of_B - 1;bound = 2;}compare_from_A = A[end_1];compare_from_B = B[end_2];if (compare_from_A > compare_from_B) {if (2 == bound) {/*stop*/temp = start_1 + kth_smallest - (end_2 - start_1 + 1) - 1;  fprintf(stdout, "median : %d\n", A[temp]);return;}start_2 = end_2 + 1;} else {start_1 = end_1 + 1;if (1 == bound) {/*stop*/temp = start_2 + kth_smallest - (end_1 - start_1 + 1) - 1;  fprintf(stdout, "median : %d\n", B[temp]);return;}}kth_smallest = (kth_smallest + 1) / 2;}//find_median_number(A, start_1, B, start_2, (kth_smallest + 1) / 2, size_of_A, size_of_B);}int main() {int a[] = {1, 2, 3, 4, 5 , 6, 7, 8, 9, 10};int b[] = {1, 2, 3 };int kth_smallest = (sizeof(a) / sizeof(int) + sizeof(b) / sizeof(int) + 1) / 2;find_median_number(a, 0, b, 0, kth_smallest, sizeof(a) / sizeof(int), sizeof(b) / sizeof(int));return 0;}

一些细节的问题还是要考虑清楚的。