﻿ 【一天一道LeetCode】#232. Implement Queue using Stacks - 鸿网互联

# 一天一道LeetCode

## （一）题目

Implement the following operations of a queue using stacks.

• push(x) – Push element x to the back of queue.

• pop() – Removes the element from in front of queue.

• peek() – Get the front element.

• empty() – Return whether the queue is empty.

Notes:
+ You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid.
+ Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
+ You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue)

## （二）解题

push：和栈一样，直接push到一个栈stack1中
pop：有一个辅助栈就可以将stack1中的数依次push到stack2中，这样stack2的top就是最先进来的数，直接pop即可
peek：去队首的数，如果stack2不为空的话，直接返回stack2.pop即可，如果stack2为空，则先把stack1的数push过来，再去top元素
empty：如果stack1和stack2均为空，就直接返回true，否则返回false

``````class Queue {
public:
// Push element x to the back of queue.
void push(int x) {
st1.push(x);
}

// Removes the element from in front of queue.
void pop(void) {
if(empty()) return;
if(st2.empty())
{
while(!st1.empty()){//为空的话要把stack1的元素push进来
st2.push(st1.top());
st1.pop();
}
}
st2.pop();
}

// Get the front element.
int peek(void) {
if(st2.empty()){//为空的话要把stack1的元素push进来
while(!st1.empty()){
st2.push(st1.top());
st1.pop();
}
}
return st2.top();
}

// Return whether the queue is empty.
bool empty(void) {
if(st1.empty()&&st2.empty()) return true;
else return false;
}
private:
stack<int> st1;
stack<int> st2;
};``````

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