LeetCode 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 top, peek/pop from top, size, 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).

思路分析：這題和幾年前寫過的一篇面經博文是相同的題目： 面試題研究 用兩個棧模擬實現隊列。基本做法很簡單，用兩個棧就可以模擬一個隊列，基本思路是兩次後進先出 = 先進先出，元素入隊列總是入A棧，元素出隊列如果B棧不爲空直接彈出B棧頭元素；如果B棧爲空就把A棧元素出棧全部壓入B棧，再彈出B棧頭，這樣就模擬出了一個隊列。核心就是保證每個元素出棧時都經過了A，B兩個棧，這樣就實現了兩次後進先出=先進先出。

AC Code

class MyQueue {
    // Push element x to the back of queue.
    Stack<Integer> stackA = new Stack<Integer>();
    Stack<Integer> stackB = new Stack<Integer>();
    
    
    public void push(int x) {
        stackA.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        if(!stackB.isEmpty()) {
            stackB.pop();
        } else {
            while(!stackA.isEmpty()){
                stackB.push(stackA.pop());
            }
            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();
        }
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return (stackA.isEmpty() && stackB.isEmpty());
    }
}