#include<iostream>
#include<vector>
using namespace std;
vector <int> removeFirstOrder(const vector<int>& orders)
{
return vector<int>(++orders.begin() , orders.end());
}
bool isFirstComeFirstServed(const vector<int>& takeOutOrders,
const vector<int>& dineInOrders,
const vector<int>& servedOrders)
{
//base case
if(servedOrders.empty())
{
return true;
}
if(!takeOutOrders.empty() && takeOutOrders[0]==servedOrders[0])
{
return isFirstComeFirstServed(removeFirstOrder(takeOutOrders),
dineInOrders,removeFirstOrder(servedOrders));
}
else if(!dineInOrders.empty() && dineInOrders[0]==servedOrders[0])
{
return isFirstComeFirstServed(takeOutOrders, removeFirstOrder(takeOutOrders),
removeFirstOrder(servedOrders));
}
else
{
return false;
}
}
int main()
{
vector<int> takeOutOrders{17,8,4};
vector<int> dineInOrders{12,19,2};
vector<int> servedOrders{17,8,12,19,24,2};
isFirstComeFirstServed(takeOutOrders,dineInOrders,servedOrders);
return 0;
}
我的疑问是这个程序的作者在这里说它有O(n^2)的时间复杂度和O(n^2)的空间复杂度,我同意这个程序的时间复杂度,因为isFirstComeFirstServed函数会被调用n次,也就是servedOrders Vector的大小,对吗?
我同意这个程序的时间复杂度,因为isFirstComeFirstServed函数会被调用n次,也就是servedOrderers向量的大小,而removeFirstOrder会在isFirstComeFirstServed的第一次函数调用中被调用n次,在isFirstComeFirstServed的第二次函数调用中被调用n-1次,以此类推,直到servedOrder向量中没有元素了为止。
但我的疑惑是,它的空间复杂度如何能达到O(n^2)呢? 有人能帮我把它可视化吗?