栈的push、pop序列 【微软面试100题 第二十九题】

// 《剑指Offer——名企面试官精讲典型编程题》代码
// 著作权所有者：何海涛

#include <iostream>
#include <stack>

using namespace std;

bool IsPopOrder(const int* pPush, const int* pPop, int nLength)
{
bool bPossible = false;

if(pPush != NULL && pPop != NULL && nLength > 0)
{
const int* pNextPush = pPush;
const int* pNextPop = pPop;

std::stack<int> stackData;

while(pNextPop - pPop < nLength)
{
// 当辅助栈的栈顶元素不是要弹出的元素
// 先压入一些数字入栈
while(stackData.empty() || stackData.top() != *pNextPop)
{
// 如果所有数字都压入辅助栈了，退出循环
if(pNextPush - pPush == nLength)
break;

stackData.push(*pNextPush);

pNextPush ++;
}

if(stackData.top() != *pNextPop)
break;

stackData.pop();
pNextPop ++;
}

if(stackData.empty() && pNextPop - pPop == nLength)
bPossible = true;
}

return bPossible;
}

// ====================测试代码====================
void Test(char* testName, const int* pPush, const int* pPop, int nLength, bool expected)
{
if(testName != NULL)
printf("%s begins: ", testName);

if(IsPopOrder(pPush, pPop, nLength) == expected)
printf("Passed.\n");
else
printf("failed.\n");
}

void Test1()
{
const int nLength = 5;
int push[nLength] = {1, 2, 3, 4, 5};
int pop[nLength] = {4, 5, 3, 2, 1};

Test("Test1", push, pop, nLength, true);
}

void Test2()
{
const int nLength = 5;
int push[nLength] = {1, 2, 3, 4, 5};
int pop[nLength] = {4,3,5,1,2};

Test("Test2", push, pop, nLength, false);
}

// push和pop序列只有一个数字
void Test5()
{
const int nLength = 1;
int push[nLength] = {1};
int pop[nLength] = {2};

Test("Test5", push, pop, nLength, false);
}

void Test6()
{
const int nLength = 1;
int push[nLength] = {1};
int pop[nLength] = {1};

Test("Test6", push, pop, nLength, true);
}

void Test7()
{
Test("Test7", NULL, NULL, 0, false);
}

int main(void)
{
Test1();
Test2();
Test5();
Test6();
Test7();

return 0;
}