[leetcode] Implement Queue using Stacks

题目链接在此。

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).

自认为是比较经典的一道题：用栈的结构和操作去实现队列的结构和操作。

很容易想到用2个栈，第一个栈的顶部是队列的尾，第二个栈的顶部是队列的头。

class Queue {public:// Push element x to the back of queue.void push(int x) {a.push(x);}// Removes the element from in front of queue.void pop(void) {if (empty()) {cout << "The queue is empty!\n";return;}if (!b.empty())b.pop();else {while (!a.empty()) {int tmp = a.top();a.pop();b.push(tmp);}b.pop();}}// Get the front element.int peek(void) {if (empty()) {cout << "The queue is empty!\n";return -1;}if (!b.empty())return b.top();else {while (!a.empty()) {int tmp = a.top();a.pop();b.push(tmp);}return b.top();}}// Return whether the queue is empty.bool empty(void) {return a.empty() && b.empty();}private:stack<int> a;stack<int> b;};