c++ - 查找以前的塔数少于或等于当前的塔数

嗨，我正在尝试查找小于或等于当前塔数的先前塔数，此解决方案对于输入（NumOfTowers） 10，代码将在segfault中运行，我无法在这里看到问题，

#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。

最佳答案

改变这个

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

对此

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}您的代码尝试访问空堆栈的top()时将给出错误。

关于c++ - 查找以前的塔数少于或等于当前的塔数，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46883974/