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  • Leetcode 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).

    解题思路:

    只要实现对了push函数,后面三个直接调用队列的函数即可。这种方法的原理就是每次把新加入的数插到前头,这样队列保存的顺序和栈的顺序是相反的,它们的取出方式也是反的,那么反反得正,就是我们需要的顺序了。我们需要一个辅助队列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队列的操作即可。


     Java Code:

    import java.util.LinkedList;
    import java.util.Queue;
    
    public class ImStackUsingQue {
        class MyStack {
            private Queue<Integer> q = new LinkedList<Integer>();
            // Push element x onto stack.
            public void push(int x) {
                Queue<Integer> temp = new LinkedList<Integer>();
                while(!q.isEmpty()) {
                    temp.add(q.peek());
                    q.remove();
                }
                q.add(x);
                while(!temp.isEmpty()) {
                    q.add(temp.peek());
                    temp.remove();
                }
            }
    
            // Removes the element on top of the stack.
            public void pop() {
                q.remove();
            }
    
            // Get the top element.
            public int top() {
                return q.peek();
            }
    
            // Return whether the stack is empty.
            public boolean empty() {
                return q.isEmpty();
            }
        }
    }

    Reference:

    1. http://www.cnblogs.com/grandyang/p/4568796.html

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  • 原文地址:https://www.cnblogs.com/anne-vista/p/4793506.html
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