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笔试 - 2014创新工场涂鸦移动校园招聘测试题 及 答案

2019-07-14 11:59发布

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2014创新工场涂鸦移动校园招聘测试题 及 答案


本文地址: http://blog.csdn.net/caroline_wendy/article/details/28664807
by Spike 2014.6.5本题目仅供学术交流, 严禁用于其他目的, 答案仅供参考.
1. 合并有序链表, 给定两个升序的链表, 返回一个合并之后的升序链表.节点结构:struct Node{int val;Node *next;};要求实现的函数：Node* mergeList (Node *list_a, Node* list_b)代码:/* * test.cpp * * Created on: 2014.04.24 * Author: Spike */ /*eclipse cdt, gcc 4.8.1*/ #include #include using namespace std; struct Node{ int val; Node *next; }; Node* mergeList(Node *list_a, Node* list_b) { if (list_a == NULL) //递归的终止条件 return list_b; else if (list_b == NULL) return list_a; Node* pMergedHead = NULL; //合并后的链表 if (list_a->val < list_b->val) { pMergedHead = list_a; //指向头结点 pMergedHead->next = mergeList(list_a->next, list_b); //递归 } else { pMergedHead = list_b; pMergedHead->next = mergeList(list_a, list_b->next); } return pMergedHead; } Node* initList(const std::vector& vi) { Node* pHead = new Node; Node* pTemp = pHead; for (std::size_t i=0; ival = vi[i]; if (i != vi.size()-1) { //非尾结点 Node* pNode = new Node; pTemp->next = pNode; pTemp = pTemp->next; } } pTemp->next = NULL; return pHead; } void printList(Node* L) { Node* pTemp = L; while (pTemp->next != NULL) { std::cout << pTemp->val << " "; pTemp = pTemp->next; } std::cout << pTemp->val << " "; //打印最后一个值 std::cout << std::endl; } int main(void) { std::vector via = {1, 2, 3, 4, 5, 13}; std::vector vib = {2, 4, 5, 7, 9, 11}; Node* list_a = initList(via); Node* list_b = initList(vib); std::cout << "list_a = "; printList(list_a); std::cout << "list_b = "; printList(list_b); Node* list_merge = mergeList(list_a, list_b); std::cout << "list_merge = "; printList(list_merge); return 0; }
输出:list_a = 1 2 3 4 5 13 list_b = 2 4 5 7 9 11 list_merge = 1 2 2 3 4 4 5 5 7 9 11 13

2. 大整数乘法, 给定两个长度不超过10000的整数, 返回乘法的结果.char* multi(char* number_a, char* number_b)代码:/* * test.cpp * * Created on: 2014.04.24 * Author: Spike */ /*eclipse cdt, gcc 4.8.1*/ #include #include using namespace std; char* multi(char* number_a, char* number_b) { int len_a = strlen(number_a); //计算长度 int len_b = strlen(number_b); int* num_arr = new int[len_a+len_b]; memset(num_arr, 0, sizeof(int)*(len_a+len_b)); //置0 for (int i=len_a-1; i>=0; --i) { //计算每一位 for (int j=len_b-1; j>=0; --j) { num_arr[i+j+1] += (number_b[j]-'0')*(number_a[i]-'0'); } } for (int i=len_a+len_b-1; i>=0; --i) { //进位 if (num_arr[i] >= 10) { num_arr[i-1] += num_arr[i]/10; num_arr[i] %= 10; } } char* result = new char[len_a+len_b+1]; for( int i=0; i<(len_a+len_b); ++i){ result[i] = (char)(((int)'0')+num_arr[i]); } result[len_a+len_b] = '

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