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out 栈相当于一个缓冲区域，把一部分入队的放到这个缓冲区域

class MyQueue {        Stack
   
     input;    Stack
    
      output;    /** Initialize your data structure here. */    public MyQueue() {        input = new Stack();        output = new Stack();                    }        /** Push element x to the back of queue. */    public void push(int x) {        input.push(x);    }        /** Removes the element from in front of queue and returns that element. */    public int pop() {        if(!output.isEmpty()) {           return output.pop();        }else {            while(!input.isEmpty()) {                 output.push(input.pop());            }            return output.pop();                    }            }        /** Get the front element. */    public int peek() {                if(!output.isEmpty()) {            return output.peek();        }else {                        while(!input.isEmpty()) {                output.push(input.pop());            }                        return output.peek();                                                }                            }        /** Returns whether the queue is empty. */    public boolean empty() {        return output.isEmpty()&&input.isEmpty();    }}/** * Your MyQueue object will be instantiated and called as such: * MyQueue obj = new MyQueue(); * obj.push(x); * int param_2 = obj.pop(); * int param_3 = obj.peek(); * boolean param_4 = obj.empty(); */下面是双队列实现
    
   

队列实现栈

我对这道题的理解是，准备两个队列，我分别命名为 input,队列和output 队列

output队列的第一位始终是最后一个进队列的，因此，每一次弹栈，都是弹出最后那个进队列的，这样就可以模拟栈的特性了 具体代码如下 `

class MyStack {Queue
   
     input;Queue
    
      output;/** Initialize your data structure here. */public MyStack() {        input = new LinkedList();    output = new LinkedList();                }/** Push element x onto stack. */public void push(int x) {        if(output.size()==0) {        output.offer(x);    }else {        while(output.size()!=0) {            input.offer(output.poll());//清空 output 队列        }            output.offer(x);           //进队的放在output队列的第一位        while(input.size()!=0) {            output.offer(input.poll());//重改output 队列        }    }                }/** Removes the element on top of the stack and returns that element. */public int pop() {        if(output.peek()==null) {        return 0;    }    return output.poll();    }/** Get the top element. */public int top() {        if(output.peek()==null) {        return 0;    }    return output.peek();}/** Returns whether the stack is empty. */public boolean empty() {    return output.peek()==null;}}
    
   

作者：deep-dark

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