LeetCode 232. 用栈实现队列

题目描述：

解法一（自然解法使用两个栈 入队 - O(1)， 出队 - O(n)）：

class MyQueue {
    stack<int> stk1;
    stack<int> stk2;
    int Front;
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        if(stk1.size()==0) Front=x;
        stk1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        if(!stk1.empty()){
            int res=Front;
            while(stk1.size()>1){
                stk2.push(stk1.top());
                stk1.pop();
            }
            stk1.pop();
            Front=stk2.empty()? -1:stk2.top();
            while(!stk2.empty()){
                stk1.push(stk2.top());
                stk2.pop();
            }
            return res;    
        }
        else return -1;
    }
    
    /** Get the front element. */
    int peek() {
        return Front;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return stk1.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();
 */

解法二（改进版使用两个栈 入队 - O(1)，出队 - 摊还复杂度 O(1)）：

class MyQueue {
    stack<int> stk1;
    stack<int> stk2;
    int Front;
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        if(stk1.empty()) Front=x;
        stk1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        if(!stk2.empty()){
            int res=stk2.top();
            stk2.pop();
            return res;
        }
        else{
            while(!stk1.empty()){
                stk2.push(stk1.top());
                stk1.pop();
            }
            int res=stk2.top();
            stk2.pop();
            return res;
        }
    }
    
    /** Get the front element. */
    int peek() {
        if(!stk2.empty())
            return stk2.top();
        else return Front;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return stk2.empty()&&stk1.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();
 */

学习到一个概念：摊还复杂度。

解法三使用两个栈 入队 - O(n)， 出队 - O(1)：

class MyQueue {
    stack<int> stk1;
    stack<int> stk2;
    int Front;
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        if(stk1.empty()) Front=x;
        while(!stk1.empty()){
            stk2.push(stk1.top());
            stk1.pop();
        }
        stk2.push(x);
        while(!stk2.empty()){
            stk1.push(stk2.top());
            stk2.pop();
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int res=Front;
        stk1.pop();
        Front=stk1.empty()? -1:stk1.top();
        return res;
    }
    
    /** Get the front element. */
    int peek() {
        return Front;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return stk1.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();
 */

参考：https://leetcode-cn.com/problems/implement-queue-using-stacks/solution/yong-zhan-shi-xian-dui-lie-by-leetcode/