[LeetCode] Implement Queue using Stacks 用栈来实现队列

标签：

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).

这道题让我们用栈来实现队列，之前我们做过一道相反的题目Implement Stack using Queues 用队列来实现栈，是用队列来实现栈。这道题颠倒了个顺序，起始并没有太大的区别，栈和队列的核心不同点就是栈是先进后出，而队列是先进先出，那么我们要用栈的先进后出的特性来模拟出队列的先进先出。那么怎么做呢，其实很简单，只要我们在插入元素的时候每次都都从前面插入即可，比如如果一个队列是1,2,3,4，那么我们在栈中保存为4,3,2,1，那么返回栈顶元素1，也就是队列的首元素，则问题迎刃而解。所以此题的难度是push函数，我们需要一个辅助栈tmp，把s的元素也逆着顺序存入tmp中，此时加入新元素x，再把tmp中的元素存回来，这样就是我们要的顺序了，其他三个操作也就直接调用栈的操作即可，参见代码如下：

class Queue {
public:
    // Push element x to the back of queue.
    void push(int x) {
        stack<int> tmp;
        while (!s.empty()) {
            tmp.push(s.top());
            s.pop();
        }
        s.push(x);
        while (!tmp.empty()) {
            s.push(tmp.top());
            tmp.pop();
        }
    }

    // Removes the element from in front of queue.
    void pop(void) {
        s.pop();
    }

    // Get the front element.
    int peek(void) {
        return s.top();
    }

    // Return whether the queue is empty.
    bool empty(void) {
        return s.empty();
    }

private:
    stack<int> s;
};

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Implement Queue using Stacks 用栈来实现队列

标签：

原文地址：http://www.cnblogs.com/grandyang/p/4626238.html