[解题报告]leetcode-232 用栈实现队列

题目，难度为简单

大意是使用两个栈来实现队列的操作，只能用基础栈操作。要求摊还复杂度O(1)，即N个动作复杂度O(n)

题解一：脑抽解法

摊还O(N)

class MyQueue {
private:
    stack<int> instack;
    stack<int> outstack;
public:
    /** Initialize your data structure here. */
    MyQueue() {

    }

    /** Push element x to the back of queue. */
    void push(int x) {
        while(!outstack.empty()){//若出栈为空，则全部赋值
            instack.push(outstack.top());
            outstack.pop();
        }
        instack.push(x);
    }

    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        while(!instack.empty()){
            outstack.push(instack.top());
            instack.pop();
        }
        int ele = outstack.top();
        outstack.pop();
        return ele;
    }

    /** Get the front element. */
    int peek() {
        while(!instack.empty()){
            outstack.push(instack.top());
            instack.pop();
        }
        int ele = outstack.top();
        return ele;
    }

    /** Returns whether the queue is empty. */
    bool empty() {
        return instack.empty()&&outstack.empty();
    }

};

/**
 * 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();
 */

虽然在执行中击败了100%，但这其实并不满足题目要求，因为每一次切换输入输出的时候都会清空一次栈。

当时的想法就是认为一个栈必须是空的，这样倒腾才能够维持内部数据的顺序。但其实并不一定要这样

解法二：正确解法

与解法一的区别有二，第一是删除了入队操作时要把输出栈导入到输入栈的操作；第二是在出队时，输入栈往输出栈运送元素处增加了一个判定。之前有点死脑经，可能是学AVL学上头了，一直想着说要维持数据结构。其实根本不用维持，因为输出栈在完全输出之前，是可以不用输入的，这就形成了一个局部有序的情况。

class MyQueue {
private:
    stack<int> instack;
    stack<int> outstack;
public:
    /** Initialize your data structure here. */
    MyQueue() {

    }

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

    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        if(outstack.empty())
        {
            while(!instack.empty()){
                outstack.push(instack.top());
                instack.pop();
            }
        }
        int ele = outstack.top();
        outstack.pop();
        return ele;
    }

    /** Get the front element. */
    int peek() {
        if(outstack.empty())
        {
            while(!instack.empty()){
                outstack.push(instack.top());
                instack.pop();
            }
        }
        int ele = outstack.top();
        return ele;
    }

    /** Returns whether the queue is empty. */
    bool empty() {
        return instack.empty()&&outstack.empty();
    }

};

/**
 * 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();
 */

wtysos11 / blogWiki

[解题报告]leetcode-232 用栈实现队列 #100

题解一：脑抽解法

解法二：正确解法