• 【LeetCode】225. Implement Stack using Queues


    Implement Stack using Queues

    Implement the following operations of a stack using queues.

    • push(x) -- Push element x onto stack.
    • pop() -- Removes the element on top of the stack.
    • top() -- Get the top element.
    • empty() -- Return whether the stack is empty.

    Notes:

    • You must use only standard operations of a queue -- which means only push to backpeek/pop from frontsize, and is empty operations are valid.
    • Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
    • You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

    Update (2015-06-11):
    The class name of the Java function had been updated to MyStack instead of Stack.

    Credits:
    Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

    两个队列实现栈:

    一个队列存放当前所有元素,另一个队列为空。

    push: 元素进非空队列,并更新top值为当前进队元素。

    pop: 将非空队列的值,除了最后一个(待出队),其余的装入空队列。注意更新top值为最后第二个进队元素。

    top: 直接返回实时更新的top值

    empty: 判断两个栈是否均为空

    class Stack {
    public:
        queue<int> q1;
        queue<int> q2;
        int n;
        
        // Push element x onto stack.
        void push(int x) {
            if(q2.empty())
                q1.push(x);
            else
                q2.push(x);
            n = x;
        }
    
        // Removes the element on top of the stack.
        void pop() {
            if(!q1.empty())
            {
                while(q1.size() > 1)
                {
                    int front = q1.front();
                    if(q1.size() == 2)
                        n = front;
                    q1.pop();
                    q2.push(front);
                }
                q1.pop();
            }
            else
            {
                while(q2.size() > 1)
                {
                    int front = q2.front();
                    if(q2.size() == 2)
                        n = front;
                    q2.pop();
                    q1.push(front);
                }
                q2.pop();
            }
        }
    
        // Get the top element.
        int top() {
            return n;
        }
    
        // Return whether the stack is empty.
        bool empty() {
            return q1.empty() && q2.empty();
        }
    };

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4640689.html
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