亚马逊在线笔试题目

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。

char* calculateOperationSequence(int * originalArray, int * resultArray, int length)

{

if (NULL == originalArray || NULL == resultArray || length <= 0)

return NULL;

//使用一个栈模拟入栈和出栈操作就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;

}

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