leetcode [#232]

目录

• 1. 题目
• 2. 解决方案
• 3. 注意事项

题目

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.

Note

• 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).

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解决方案

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

    // Push element x to the back of queue.
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        while (!stack1.empty()){
            stack2.push(stack1.pop());
        }
        stack2.pop();
        while (!stack2.empty()){
            stack1.push(stack2.pop());
        }
    }

    // Get the front element.
    public int peek() {
        while (!stack1.empty()){
            stack2.push(stack1.pop());
        }
        int res = stack2.peek();
        while (!stack2.empty()){
            stack1.push(stack2.pop());
        }
        return res;
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.empty() ? true : false;
    }
}

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注意事项

1. 使用两个栈来实现队列。
2. 思路是，对于push()操作，直接向第一个栈push即可；对于pop()操作，将第一个栈所有元素弹出并压入第二个栈，弹出第二个栈的栈顶元素，再把剩下的元素依次弹出并压回第一个栈；对于peek()操作，将第一个栈所有元素弹出并压入第二个栈，查看第二个栈的栈顶元素（只是查看），之后将第二个栈的所有元素依次弹出并压回第一个栈。