Question

您好我试图找到小于或等于当前塔的数量，此解决方案适用于输入（NumOfTowers）＆lt; = 10，但对于NumOfTowers＆gt; 10代码进入segfault，我在这里看不到问题，

#include <iostream>
#include <stack>
using namespace std;

int main()
{
    std::stack<int> Towers;    //for storing the Height of the towers
    std::stack<int> TempTowers; // Temrory buffer stack
    std::stack<int> CountTowers; //for storing the count of all previous  
                                   towers less than the current one
    unsigned int NumTestCases,NumOfTowers,i,j,count=1,temp,temp_height;
    cin>>NumTestCases;
    cin>>NumOfTowers;


    while(NumTestCases){

        while(!Towers.empty()){
            Towers.pop();
        }
       for(i=0;i<NumOfTowers;i++){
           cin>>temp;
           Towers.push(temp);
       }
       for(i=0;i<NumOfTowers-1;i++){
           count=1;
           temp_height=Towers.top();
           Towers.pop();
           temp=Towers.top();

           while(temp<temp_height){
               count++;
               TempTowers.push(temp);
               Towers.pop();
               temp=Towers.top();
           }

            CountTowers.push(count);

            while(!TempTowers.empty()){
                temp=TempTowers.top();
                TempTowers.pop();
                Towers.push(temp);
           }

       }       
       NumTestCases--;
       cout<<"1"<<" ";
       while(!CountTowers.empty()){
           cout<<CountTowers.top()<<" ";
           CountTowers.pop();
       }
       cout<<"\n";
    }

}

任何帮助都会很棒。

Answer 1

更改此

while(temp<temp_height){
               count++;
               TempTowers.push(temp);
               Towers.pop();      // Popped the last element
               temp=Towers.top(); // no more elements left
           }

到这个

while(!Towers.empty() && Towers.top() < temp_height)
{
  ++count;
  TempTowers.push(Towers.top());
  Towers.pop();
}

问题不在于输入的长度，而在于输入的顺序。如果I/p = {1, 2, 3, 4}您的代码在尝试访问空堆栈的top()时会出错。

Answer 2

您正在空堆栈上调用temp=Towers.top();，此处

    while(temp<temp_height){        
       count++;
       TempTowers.push(temp);
       Towers.pop();
       temp=Towers.top();
    }

将此行更改为以下两行：

    if (Towers.size() > 0) temp = Towers.top();
    else break;

找到小于或等于当前塔的先前塔的数量

2 个答案: