LeetCode 232：Implement Queue using Stacks



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).

//题目描述://使用栈实现队列的下列操作：//push(x) --将元素x加至队列尾部//pop( ) --从队列头部移除元素//peek( ) --获取队头元素//empty( ) --返回队列是否为空//注意：你只可以使用栈的标准操作，这意味着只有push to top（压栈）, peek / pop from top（取栈顶 / 弹栈顶），//以及empty（判断是否为空）是允许的，取决于你的语言，stack可能没有被内建支持。你可以使用list（列表）或者deque（双端队列）来模拟，//确保只使用栈的标准操作即可，你可以假设所有的操作都是有效的（例如，不会对一个空的队列执行pop或者peek操作）//解题方法：用两个栈就可以模拟一个队列，基本思路是两次后进先出 = 先进先出，//元素入队列总是入in栈，元素出队列如果out栈不为空直接弹出out栈头元素；//如果out栈为空就把in栈元素出栈全部压入out栈，再弹出out栈头，这样就模拟出了一个队列。//核心就是保证每个元素出栈时都经过了in，out两个栈，这样就实现了两次后进先出=先进先出。class Queue {public:stack<int> in;stack<int> out;void move(){   //将in栈内的所有元素移动到out栈while (!in.empty()){int x = in.top();in.pop();out.push(x);}}// Push element x to the back of queue.void push(int x) {in.push(x);}// Removes the element from in front of queue.void pop(void) {if (out.empty()){move();}if (!out.empty()){out.pop();}}// Get the front element.int peek(void) {if (out.empty()){move();}if (!out.empty()){return out.top();}}// Return whether the queue is empty.bool empty(void) {return in.empty() && out.empty();}};