发布于 2014-12-29 09:55:49 | 513 次阅读 | 评论: 0 | 来源: 网友投递
亚马逊(Amazon)
亚马逊公司（Amazon，简称亚马逊；NASDAQ：AMZN），是美国最大的一家网络电子商务公司，位于华盛顿州的西雅图。是网络上最早开始经营电子商务的公司之一，亚马逊成立于1995年，一开始只经营网络的书籍销售业务，现在则扩及了范围相当广的其他产品，已成为全球商品品种最多的网上零售商和全球第二大互联网企业，在公司名下，也包括了AlexaInternet、a9、lab126、和互联网电影数据库（Internet Movie Database，IMDB）等子公司。
本文为大家提供的是一份亚马逊在线笔试题目-技术开发类岗位，e文题，感兴趣的同学参考下。
亚马逊公司（Amazon，简称亚马逊；NASDAQ：AMZN），是美国最大的一家网络电子商务公司，位于华盛顿州的西雅图。是网络上最早开始经营电子商务的公司之一，亚马逊成立于1995年，一开始只经营网络的书籍销售业务，现在则扩及了范围相当广的其他产品，已成为全球商品品种最多的网上零售商和全球第二大互联网企业，在公司名下，也包括了AlexaInternet、a9、lab126、和互联网电影数据库（Internet Movie Database，IMDB）等子公司。
2小时，两个题目。在线编程，英文题目。当时没做好，完了自己把他们完成了。答案是我自己写的，自己测试没问题，若有错误请指正。
Question 1 / 2
Question:
We have an array representing customer’s shopping records.
For example, it’s an array like this:
custA, item1,
custB, item1,
custA, item2,
custB, item3,
custC, item1,
custC, item3,
custD, item2,
This array indicates that customer A bought item 1, customer B bought item 1, customer A bought item 2, customer B bought
item 3, etc..
For a given item X and shopping records array, write code to find out what else (item Y) was bought mostly by the customers
who bought item X.
For example, in above example, if X is item 1 then Y should be item 3.
Rules:
1. One customer can only buy one item once.
2. The mostly brought item should not be item X.
3. If no customer brought item X, then return “None”
4. If all the customers who brought item X only brought item X, then return “None”
5. The first line of input is the item X. The second line of input is the shopping record array, this shopping record array is
split by space.
6. If there are many other mostly brought items which have equally brought times, then return any one of those items.
Examples:
Input1:
item1
custA item1 custB item1 custA item2 custB item3 custC item1 custC item3 custD item2
Output1:
item3
Input2:
item2
custA item1 custB item1 custC item1 custA item2 custB item3 custA item3
Output2:
item1
(The output2 can be item3 too)
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
#include
#include
#include
#include
#include
#include
#include
using namespace std;
char* findMostlyBroughtItem(char* shippingRecordArray[], int length, char* givenItem);
inline bool isSpace(char x){
return x == ' ' || x == 'r' || x == 'n' || x == 'f' || x == 'b' || x == 't';
}
char * rightTrim(char *str){
int len = strlen(str);
while(--len>=0){
if(isSpace(str[len])){
str[len] = '�';
}else{
break;
}
}
return str;
}
char * getInputLine(char *buffer, int length){
if(fgets(buffer,length, stdin)==NULL){
return NULL;
}
rightTrim(buffer);
if(strlen(buffer)<=0){
return NULL;
}
return buffer;
}
int splitAndConvert(char* strings,char* array[]){
char*tokenPtr = strtok(strings," ");
int i=0;
while(tokenPtr!=NULL){
array[i] = tokenPtr;
i++;
tokenPtr=strtok(NULL," ");
}
return i;
}
int main()
{
char givenItem[1000] = {0} ;
while(getInputLine(givenItem, 1000)){
char line[1000];
getInputLine(line, 1000);
char* shoppingRecordArray[1000] = {0};
int length = splitAndConvert(line,shoppingRecordArray);
if(length==0){
break;
}
char * item = findMostlyBroughtItem(shoppingRecordArray, length, givenItem);
if (NULL != item)
{ // 原来系统提供的代码。这里没有NULL判断
cout<
free(item); // 自己加的
}
}
return 0;
}
void
print(pair
cout << p.first << p.second << endl;
}
//your code is here
//下面才是让写代码的地方，其他的系统已经自动给出。主函数，只有一点点修改。
char* findMostlyBroughtItem(char* shoppingRecordArray[], int length, char* givenItem)
{
if (NULL == shoppingRecordArray || NULL == givenItem)
return NULL;
string obj_item(givenItem);
// 将用户信息 与 购买商品信息 存入multimap record
multimap
for (int i = 0; i < length; i += 2)
{
string customer(shoppingRecordArray[i]);
string item(shoppingRecordArray[i+1]);
record.insert(pair
}
// 提取出购买了obj_item商品的客户名称集合 customers
set customers;
for (map
{
if (0 == (*it).second.compare(obj_item))
{
customers.insert((*it).first);
}
}
// 遍历购买记录 multimap record
// 若客户名称 在 集合set customers 存在，则将商品插入map result
map
for (map
{
for (set::iterator ic = customers.begin(); ic != customers.end(); ic++)
{
if (0 == (*it).first.compare(*ic))
{
/*
if (result.end() != result.find((*it).second))
{
result[(*it).second] += 1;
}
else
result.insert(pair
*/
result[(*it).second] += 1;
break;
}
}
}
pair
// 遍历map result, 寻找最大，而非obj_item的商品名称
for (map
{
if (0 == (*it).first.compare(obj_item))
continue;
if ((*it).second > top.second)
top = make_pair((*it).first, (*it).second);
}
//cout << "Top: " << top.first << "t" << top.second << endl;
char *p = (char *)malloc(top.first.length() + 1);
if (NULL != p)
{
strcpy(p, top.first.c_str());
return p;
}
return NULL;
}