概要：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;}elseresult.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&qu
亚马逊在线笔试题目,标签：笔试大全,http://www.88haoxue.com{
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;
} www.88haoxue.com
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:
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)
{ // 原来系统提供的代码。这里没有NULL判断
cout<< operationSequence <
free(operationSequence); // 自己加的
}
else
cout<< "None" <
}
return 0;
}
//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。
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