解题思路
栈,一个口,先进后出;队列,两个口,先进先出;
两个栈就有两个口,一个当入口,另一个当出口
当stack2为空,将stack1元素push到stack2,再pop stack2 ; 当stack2不为空,直接pop stack2
code
#include <stack>
class Solution
{
public:void push(int node) {stack1.push(node);}int pop() {if(stack2.empty() && stack1.empty() {throw "queue is empty"}if(stack2.empty()) {while(!stack1.empty()) {stack2.push(stack1.top());stack1.pop();}}int value = stack2.top();stack2.pop(); return value;}
private:stack<int> stack1;stack<int> stack2;
}