PAT A1051. Pop Sequence (25)

题目描述

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, …, N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

输入

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

输出

For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not.

样例输入

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

样例输出

YES
NO
NO
YES
NO

解题思路

判断合法出栈队列,类似Leetcode

  1. 创建等长队列并将待测序列push入队列
  2. 分别将1-N个数push入栈,每次入栈后分别与队前元素进行比较,若相等则同时弹出栈顶和队前元素
  3. 若最后队列为空就是合法序列,反之就是不合法
  4. 和Leetcode区别是给定了栈最大容量,因此push前判断是否栈满,若栈满直接说明序列不合法

AC代码

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#include<iostream>
#include<queue>
#include<stack>
using namespace std;
int main()
{
int M,N,K;
while(cin>>M>>N>>K)
{
for(int i=0; i<K; i++)
{
std::queue<int> q = std::queue<int>();
for(int j=1; j<=N; j++)
{
int tmp;
cin>>tmp;
q.push(tmp);
}
std::stack<int> s = std::stack<int>();
bool flag = 1;
for(int j=1; j<=N; j++)
{
if(s.size()<M)
{
s.push(j);
}
else
{
flag=0;
}
while(!s.empty() && s.top()==q.front())
{
s.pop();
q.pop();
}
}
if(!q.empty())
{
flag=0;
}
if(flag)
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
}
}
return 0;
}
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