LeetCode 232. Implement Queue using Stacks

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 toppeek/pop from topsize, 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).

题意:使用栈实现队列的如下操作:
push(x):将x加入队尾
pop():移除队首元素
peek():取队首元素
empty():判断队列是否为空
Notes:
只能使用栈的基本操作:push, pop ,peek, size, isEmpty


思路:使用两个栈。
元素出队时,先将栈stack1中除栈顶元素外的所有元素出栈,存入stack2中;移除栈顶元素(即队尾元素)后,再将stack2中的元素放入stack1中

class MyQueue {
    Stack<Integer> stack1 = new Stack<>();
    Stack<Integer> stack2 = new Stack<>();
    private int top;

    /** Initialize your data structure here. */
    public MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        if(stack1.isEmpty())//stack1.size() == 0
            top = x;
        stack1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        while(stack1.size() > 1){
            top = stack1.pop();
            stack2.push(top);
        }
        int x = stack1.pop();
        while(stack2.size() > 0){
            int i = stack2.pop();
            stack1.push(i);
        }
        return x;
    }
    
    /** Get the front element. */
    public int peek() {
        return top;
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        return stack1.isEmpty();
    }
}

 

posted @ 2018-03-14 22:02  zeroingToOne  阅读(150)  评论(0编辑  收藏  举报