2013亞馬遜在線(xiàn)筆試題目
2小時(shí)，兩個(gè)題目。在線(xiàn)編程，英文題目。當(dāng)時(shí)沒(méi)做好，完了自己把他們完成了。答案是我自己寫(xiě)的，自己測(cè)試沒(méi)問(wèn)題，若有錯(cuò)誤請(qǐng)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] = '\0';
}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)
{ // 原來(lái)系統(tǒng)提供的代碼。這里沒(méi)有NULL判斷
cout<
free(item); // 自己加的
}
}
return 0;
}
void
print(pair
cout << p.first << p.second << endl;
}
//your code is here
//下面才是讓寫(xiě)代碼的地方，其他的系統(tǒng)已經(jīng)自動(dòng)給出。主函數(shù)，只有一點(diǎn)點(diǎn)修改。
char* findMostlyBroughtItem(char* shoppingRecordArray[], int length, char* givenItem)
{
if (NULL == shoppingRecordArray || NULL == givenItem)
return NULL;
string obj_item(givenItem);
// 將用戶(hù)信息 與 購(gòu)買(mǎi)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
}
// 提取出購(gòu)買(mǎi)了obj_item商品的客戶(hù)名稱(chēng)集合 customers
set customers;
for (map
{
if (0 == (*it).second.compare(obj_item))
{
customers.insert((*it).first);
}
}
// 遍歷購(gòu)買(mǎi)記錄 multimap record
// 若客戶(hù)名稱(chēng) 在 集合set customers 存在，則將商品插入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
// 遍歷map result, 尋找最大，而非obj_item的商品名稱(chēng)
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;
}
Question 2 / 2
Question:
As you know, two operations of Stack are push and pop. Now give you two integer arrays, one is the original array before
push and pop operations, the other one is the result array after a series of push and pop operations to the first array. Please
give the push and pop operation sequence.
For example:
If the original array is a[] = {1,2,3}, and the result array is b[] = {1,3,2}.
Then, the operation sequence is “push1|pop1|push2|push3|pop3|pop2”(operations are split by ‘|’ and no space).
Rules:
Time Remaining: 00:25:17
1. The push and pop operations deal with the original int array from left to right.
2. The input is two integer array. They are the original array and the result array. These interger array is split by space.
3. The output is the operation sequence.
4. If the original array cannot make to the result array with stack push and pop, The output should be 'None'.
5. The operation "push1" means push the first element of the original array to the stack.
6. The operation "pop1" means pop the first element of the original array from the stack, and add this element to the tail
of the result array.
7. Please don't include any space in the output string.
Sample1:
Input:
1 2 3 4
1 2 3 4
Output:
push1|pop1|push2|pop2|push3|pop3|push4|pop4
Sample2:
Input:
1 2 3 4
4 3 2 1
Output:
push1|push2|push3|push4|pop4|pop3|pop2|pop1
#include
#include
#include
#include
#include
using namespace std;
char* calculateOperationSequence(int *originalArray, int *resultArray, int length);
inline bool isSpace(char x){
return x == ' ' || x == '\r' || x == '\n' || x == '\r' || x == '\b' || x == '\t';
}
char * rightTrim(char *str){
int len = strlen(str);
while(--len>=0){
if(isSpace(str[len])){
str[len] = '\0';
}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,int *array){
char*tokenPtr = strtok(strings,",");
int i=0;
while(tokenPtr!=NULL){
array[i] = atoi(tokenPtr);
i++;
tokenPtr=strtok(NULL,",");
}
return i;
}
int main(){
char line[1000] = {0} ;
while(getInputLine(line,1000)){
int originalArray[30] = {0};
int originalArrayLength = splitAndConvert(line,originalArray);
if(originalArrayLength==0){
break;
}
getInputLine(line, 1000);
int resultArray[30] = {0};
int resultArrayLength = splitAndConvert(line,resultArray);
if(resultArrayLength==0){
break;
}
char *operationSequence = calculateOperationSequence(originalArray, resultArray, resultArrayLength);
if (NULL != operationSequence)
{ // 原來(lái)系統(tǒng)提供的.代碼。這里沒(méi)有NULL判斷
cout<< operationSequence <
free(operationSequence); // 自己加的
}
else
cout<< "None" <
}
return 0;
}
//your code is here
//下面才是讓寫(xiě)代碼的地方，其他的系統(tǒng)已經(jīng)自動(dòng)給出。主函數(shù)，只有一點(diǎn)點(diǎn)修改。
char* calculateOperationSequence(int * originalArray, int * resultArray, int length)
{
if (NULL == originalArray || NULL == resultArray || length <= 0)
return NULL;
//使用一個(gè)棧模擬入棧和出棧操作就ok了。
string str;
stack st;
int i = 0;
int j = 0;
st.push(originalArray[i]);
char tmp[5] = "\0";
str.append("push");
sprintf(tmp, "%d", originalArray[i]);
str.append(tmp);
str.append("|");
i++;
while (!st.empty())
{
if (j < length && st.top() == resultArray[j])
{
str.append("pop");
sprintf(tmp, "%d", resultArray[j]);
str.append(tmp);
str.append("|");
st.pop();
j++;
if (i < length)
{
st.push(originalArray[i]);
str.append("push");
sprintf(tmp, "%d", originalArray[i]);
str.append(tmp);
str.append("|");
i++;
}
}
else
{
if (i < length)
{
st.push(originalArray[i]);
str.append("push");
sprintf(tmp, "%d", originalArray[i]);
str.append(tmp);
str.append("|");
i++;
}
else
break;
}
}
if (!st.empty())
return NULL;
char *p = (char *)malloc(1 + str.length());
if (NULL != p)
{
strcpy(p, str.c_str());
p[str.length() - 1] = '\0';
return p;
}
return NULL;
}
【2013亞馬遜在線(xiàn)筆試題目】相關(guān)文章：
中興2013年Java開(kāi)發(fā)筆試題目及答案11-16
阿里在線(xiàn)前端筆試題01-15
美的筆試題目04-09
會(huì)計(jì)筆試題目07-03
EXCEL筆試題目06-12
惠普筆試題目08-10
伊利筆試題目07-09
夏普筆試題目07-03
奧美筆試題目06-18
LG筆試最�？嫉墓P試題目07-09