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  • LeetCode -- Implement Stacks using Queue

    Question:

    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).

    Analysis:

    问题描述:用队列模仿一个栈。

    思路:用两个队列模仿一个栈。每次要pop或者peek时,使用队列倒换一下,剩下最后一个元素单独处理。当且仅当两个队列都为空时,栈才为空。

    Answer:

    class MyStack {
        Queue<Integer> q1 = new LinkedList<Integer>();
        Queue<Integer> q2 = new LinkedList<Integer>();
        
        // Push element x onto stack.
        public void push(int x) {
            q1.offer(x);
        }
    
        // Removes the element on top of the stack.
        public void pop() {
            if(!q1.isEmpty()) {
                    while(q1.size() > 1) {
                        int i = q1.poll();
                        q2.offer(i);
                    }
                    q1.poll();
            } else {
                    while(q2.size() > 1) {
                        int i = q2.poll();
                        q1.offer(i);
                    }
                    q2.poll();
            }
        }
    
        // Get the top element.
        public int top() {
            if(!q1.isEmpty()) {
                while(q1.size() > 1) {
                    int i = q1.poll();
                    q2.offer(i);
                }
                int i = q1.poll();
                q2.offer(i);
                return i;
            } else {
                while(q2.size() > 1) {
                    int i = q2.poll();
                    q1.offer(i);
                }
                int i = q2.poll();
                q1.offer(i);
                return i;
            }
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            if(q1.size() == 0 && q2.size() == 0)
                    return true;
            return false;
        }
    }
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  • 原文地址:https://www.cnblogs.com/little-YTMM/p/4848674.html
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