232. Implement Queue using Stacks

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).

分析

只能用栈的基本操作来模拟队列的基本操作，定义两个栈a和b，其中a保存队列的正序，在进队时使用；b保存队列的倒序，在出队时使用。

class Queue {private:    stack<int> a,b;//压入时用a，弹出时用bpublic:// Push element x to the back of queue.void push(int x) {    if (b.empty()) a.push(x);//b为空时证明没有pop过，直接在a中压入x    else {        while (!b.empty()){//如果队列进行过pop操作，则b保存的是pop之后的队列，是队列倒序            a.push(b.top());//因此a在压入新元素的时候先将b压入a，在a中形成正序队列，再压入x            b.pop();        }        a.push(x);    }}// Removes the element from in front of queue.void pop(void) {    if (a.empty()) b.pop();//a为空证明没有再压入新元素，b已经是队列倒序，可以直接弹出    else{        while(!a.empty()){//否则b保存a的倒序结果，然后弹出            b.push(a.top());            a.pop();        }        b.pop();    }}// Get the front element.int peek(void) {    if (a.empty()) return b.top();//a为空则与pop一样返回b的首元素    else {        while(!a.empty()){//否则b保存a的倒序结果，然后弹出b的首元素            b.push(a.top());            a.pop();        }        return b.top();    }}// Return whether the queue is empty.bool empty(void) {    return a.empty()&&b.empty();}};