剑指Offer面试题7 & Leetcode232

剑指Offer面试题7 & Leetcode232

Implement Queue using Stacks

用两个栈实现队列

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.

解题思路

  考虑：push()时可以直接push进stackA，pop()时,如果stackA不为空，则如果想拿到队列头部元素，即stackA栈底元素，需要将stackA中的元素全部pop()出并存进stackB,此时stackB的栈顶元素即为队列头部元素，pop()即可。

  当再一次pop()时，分两种情况:第一：stackB非空，则直接pop()；第二：stackB为空stackA非空，则再次执行上述操作，将stackA的元素pop到stackB。此时如果想进行push操作，扔可以直接push进stackA，因为在队列总stackA中的元素一定是在stackB元素的后面。

Solution

public class MyQueue {    private Stack<Integer> stackA = new Stack<Integer>();    private Stack<Integer> stackB = new Stack<Integer>();    /** Initialize your data structure here. */    public MyQueue() {    }    /** Push element x to the back of queue. */    public void push(int x) {       stackA.push(x);    }    /** Removes the element from in front of queue and returns that element. */    public int pop() {        if(!stackB.isEmpty()){            return stackB.pop();        }else{            while(!stackA.isEmpty()){                stackB.push(stackA.pop());            }            return stackB.pop();        }    }    /** Get the front element. */    public int peek() {        if(!stackB.isEmpty()){            return stackB.peek();        }else{            while(!stackA.isEmpty()){                stackB.push(stackA.pop());            }            return stackB.peek();        }    }    /** Returns whether the queue is empty. */    public boolean empty() {        return stackA.isEmpty() && stackB.isEmpty();    }