232. Implement Queue using Stacks

problem description

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.

Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);  
queue.peek();  // returns 1
queue.pop();   // returns 1
queue.empty(); // returns false

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).

algorithm thought

使用stack实现一个queue，这里需要两个栈。第一次做的时候，我实现的pop操作，时间复杂度为O(n)，并且认为应该没有地方能改了。但是这次抱着试一试的心态去看了discuss，发现确实有很厉害的算法，将所有操作的平均时间复杂度降低到了O(1)

使用一个input stack和output stack。如果需要peek或者pop，直接返回output stack的元素。如果output stack为空，就将input stack中所有元素push到output中。这里确实需要O(n)的时间，但是和平均O(1)不矛盾。

一般的做法是，一直弹出，直到最后一个数，返回，然后将栈复原。但是这里直接保存在output stack中。因为，只要反转一次，就不用再复原了，直接保持状态，之后就可以不用反转。

code

class MyQueue {
public:
    /** Initialize your data structure here. */
    MyQueue() {

    }

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

    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int res=peek();
        output.pop();
        return res;
    }

    /** Get the front element. */
    int peek() {
        if(output.size()==0){       
            while(input.size()){
                output.push(input.top()),input.pop();
            }    
        }
        return output.top();
    }

    /** Returns whether the queue is empty. */
    bool empty() {
        return input.empty()&&output.empty();
    }
private:
    stack<int> input,output;
};

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue* obj = new MyQueue();
 * obj->push(x);
 * int param_2 = obj->pop();
 * int param_3 = obj->peek();
 * bool param_4 = obj->empty();
 */

algorithm analysis

由于这个算法，对于每个数字只需要一次入栈操作，和一次转移栈操作以及一次出栈操作。这几个操作是O(n)的，所以对于每个数字而言，平均时间复杂度是O(1)