LeetCode--Implement Queue using Stacks

Problem:

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.

Notes:

- You must use onlystandard 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).

Analysis:

题意：用栈实现队列

用两个堆栈实现，一个放push的数据，另一个放pop的数据，但是在pop在前需要将push栈中的数据push到pop栈中,这样就可以使pop先进先出了。

注意：

堆栈的输入元素是对象，所以需要定义Integer类型

pop时应该先pop堆栈（pop的），如果为空就要将push堆栈的数据push到pop堆栈中，然后再pop出去。

peek需要返回int类型，所以注意输出时return，不要忘了。

判断是否为空主要是判断两个栈是否同时为空。

Anwser:

class MyQueue {
// Push element x to the back of queue.
Stack<Integer> in = new Stack<>();
Stack<Integer> out = new Stack<>();
public void push(int x) {
in.push(x);
}
// Removes the element from in front of queue.
public void pop() {
if (!out.empty()) out.pop();
else {
while(!in.empty())  out.push(in.pop());
out.pop();
}
}
// Get the front element.
public int peek() {
if (!out.empty()) return out.peek();
else {
while(!in.empty()) out.push(in.pop());
return out.peek();
}
}
// Return whether the queue is empty.
public boolean empty() {
if(in.empty()&& out.empty()) return true;
else return false;
}
}