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  • 225. Implement Stack using Queues

    题目:

    Implement the following operations of a stack using queues.

    • push(x) -- Push element x onto stack.
    • pop() -- Removes the element on top of the stack.
    • top() -- Get the top element.
    • empty() -- Return whether the stack is empty.

    Notes:

    • You must use only standard operations of a queue -- which means only push to backpeek/pop from frontsize, and is empty operations are valid.
    • Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
    • You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

    Update (2015-06-11):
    The class name of the Java function had been updated to MyStack instead of Stack.

    链接: http://leetcode.com/problems/implement-stack-using-queues/

    题解:

    一开始想使用两个Queue,来回倒腾一下就可以得到结果,但这样基本每个op都是O(n)。后来看了Discuss,发现一个Queue就可以,然后只有Push()的Complexity - O(n),应该算是optimal了。不过仔细想一想其实Push最好能是O(1),因为这个用得应该最频繁。 下面做法主要就是在push的时候把当前值放在队尾,以前的值我们重新放在队尾。

    Time Complexity - Push - O(n), Pop - O(1), Peek - O(1),  isEmpty- O(1)

    class MyStack {
        // Push element x onto stack.
        Queue<Integer> q;
        
        public MyStack() {
            q = new LinkedList<>();
        }
        
        public void push(int x) {
            q.offer(x);
            
            for(int i = 0; i < q.size() - 1; i++) {
                q.offer(q.peek());
                q.poll();
            }
        }
    
        // Removes the element on top of the stack.
        public void pop() {
            q.poll();
        }
    
        // Get the top element.
        public int top() {
            return q.peek();
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            return q.size() == 0;
        }
    }

    二刷:

    使用一个queue来求解,每次push的时候把q头部poll出的数字再加回到尾部

    Java:

    Time Complexity - Push - O(n), Pop - O(1), Peek - O(1),  isEmpty- O(1)

    class MyStack {
        // Push element x onto stack.
        Queue<Integer> q = new LinkedList();
        
        public void push(int x) {
            q.offer(x);
            for (int i = 0; i < q.size() - 1; i++) {
                q.offer(q.poll());
            }
        }
    
        // Removes the element on top of the stack.
        public void pop() {
            q.poll();
        }
    
        // Get the top element.
        public int top() {
            return q.peek();
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            return q.size() == 0;
        }
    }

    三刷:

    Java:

    class MyStack {
        private Queue<Integer> q = new LinkedList<>();
        
        // Push element x onto stack.
        public void push(int x) {
            q.offer(x);
            for (int i = q.size() - 2; i >= 0; i--) {
                q.offer(q.poll());
            }
        }
    
        // Removes the element on top of the stack.
        public void pop() {
            q.poll();
        }
    
        // Get the top element.
        public int top() {
            return q.peek();
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            return q.isEmpty();
        }
    }

    Update: 不过实际运行速度比较慢啊

    Time Complexity - Push - O(n), Pop - O(1), Peek - O(1),  isEmpty- O(1)

    class MyStack {
        Queue<Integer> q = new LinkedList<>();
        // Push element x onto stack.
        public void push(int x) {
            q.offer(x);
            int len = q.size();
            while (len > 1) {
                q.offer(q.poll());
                len--;
            }
        }
    
        // Removes the element on top of the stack.
        public void pop() {
            q.poll();
        }
    
        // Get the top element.
        public int top() {
            return q.peek();
        }
    
        // Return whether the stack is empty.
        public boolean empty() {
            return q.isEmpty();
        }
    }

    Reference:

    https://leetcode.com/discuss/46975/a-simple-c-solution

    https://leetcode.com/discuss/40202/only-push-others-using-queue-combination-shared-solutions 

    https://leetcode.com/discuss/84233/solutions-about-which-utilizes-queue-others-utilize-queues

    https://leetcode.com/discuss/39839/o-1-purely-with-queues

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/4994266.html
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