C/C++迅雷公司2012笔试算法题

// 打印两个升序排列数组的交集，例如：int a[17] = {-10, -6, -1, 0, 2, 5, 7, 9, 11, 15, 17, \25, 33, 47, 60, 66, 90, 95, 102}; // 19个整数int b[4] = {-15, 1, 5, 9}; // 4个整数// 实现方式一，算法的时间复杂度为O(m + n)void intersection(int a[], int m, int b[], int n){for (int i = 0, j = 0; (i < m) && (j < n);){if ( a[i] == b[j] ){printf("%d\n", a[i]); // 打印交集++i;++j;}else if( a[i] < b[j] ){++i;}else{++j;}}}// 添加约束条件：两个数组的元素个数之间存在关系n大于m^2，请设计时间复杂度最优的算法求两者的交集。// 遍历元素个数较少的数组a，逐个元素在b中进行二分查找，时间复杂度为O(m * log(m))// 而实现方式一的时间复杂度为O(m^2)#define avg(a, b) (((a) + (b)) / 2)int bsearch(int a[], int len, int value){for (int low = 0, high = len -1, mid = avg(low, high); low <= high; mid = avg(low, high) ){if ( a[mid] == value ){return mid;}else if ( a[mid] < value ){low = mid + 1;}else{high = mid - 1;}}return mid;}void intersection(int a[], int m, int b[], int n){for (int pos = 0, i = 0; i < m; ++i){pos += bsearch(b + pos, n - pos, a[i]);if ( b[pos] == a[i] ){printf("%d\n", a[i]);}}}