办法一、暴力解法
即先将两个有序数组合并为一个有序数组，而后求得此有序数组的中位数。因而算法的外围是归并两个有序数组，能够借鉴归并排序里归并两个有序数组的办法。此法工夫复杂度为O(m+n)，因为须要遍历两个数组所有元素，空间复杂度为O(m+n)，因为须要开拓新数组存储这两个数组的所有元素。
#include <stdio.h>#include <stdlib.h>double findMedienSortedArr(int *arr1, int arr1Size, int *arr2, int arr2Size){ //merge array int size = arr1Size + arr2Size; int *arr = (int *)malloc(size * sizeof(int)); int arr1Index = 0; int arr2Index = 0; double medien; for (int index = 0; index < size; index++) { if (arr1Index < arr1Size && arr2Index < arr2Size) { if (arr1[arr1Index] <= arr2[arr2Index]) { arr[index] = arr1[arr1Index]; arr1Index++; } else { arr[index] = arr2[arr2Index]; arr2Index++; } } else if (arr1Index < arr1Size && arr2Index >= arr2Size) { arr[index] = arr1[arr1Index]; arr1Index++; } else if (arr1Index >= arr1Size && arr2Index < arr2Size) { arr[index] = arr2[arr2Index]; arr2Index++; } } // 打印合并后的数组，测试的代码，能够正文掉 //for (int j = 0; j < size; j++) { // printf("%d ", arr[j]); //} if (size % 2 == 0) { medien = (double)(arr[size / 2 - 1] + arr[size / 2]) / 2; } else { medien = arr[size / 2]; } return medien;}int main(void){ int arr1[] = {2, 4, 5, 7, 9, 29, 67}; int arr2[] = {3, 6, 8, 9, 10, 13, 18}; int size1 = sizeof(arr1) / sizeof(*arr1); int size2 = sizeof(arr2) / sizeof(*arr2); printf("arr length: %d %d\n", size1, size2); double medien = findMedienSortedArr(arr1, size1, arr2, size2); printf("\n"); for (int i = 0; i < size1; i++) { printf("%d ", arr1[i]); } printf("\n"); for (int j = 0; j < size2; j++) { printf("%d ", arr2[j]); } printf("\n"); printf("medien is %f", medien); printf("\n");}