代码随想录day10：栈与队列part01（用栈实现队列，用队列实现栈）

栈与队列part01

用栈实现队列

时间复杂度: push和empty为O(1), pop和peek为O(n)
空间复杂度: O(n)

class MyQueue {
public:
    stack<int> stIn;  // 输入栈
    stack<int> stOut; // 输出栈

    MyQueue() {

    }
    
    void push(int x) {
        stIn.push(x);
    }
    
    int pop() {
        // 只有当输出栈为空时，才从输入栈里导入全部数据
        if(stOut.empty()) {
            while(!stIn.empty()) {
                stOut.push(stIn.top());
                stIn.pop();
            }
        }
        int res = stOut.top();
        stOut.pop();
        return res;
    }
    
    int peek() {
        int res = this->pop();
        stOut.push(res); // pop函数弹出了res，所以再添加回去
        return res;
    }
    
    bool empty() {
        return stIn.empty() && stOut.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();
 */

用队列实现栈

时间复杂度: pop为O(n)，其他为O(1)
空间复杂度: O(n)

class MyStack {
public:
    queue<int> que;

    MyStack() {

    }
    
    void push(int x) {
        que.push(x);
    }
    
    int pop() {
        int size = que.size();
        size--;
        // 将队列的头部元素（除最后一个元素外），重新添加到队列的尾部
        while(size--) { 
            que.push(que.front());
            que.pop();
        }
        int res = que.front();
        que.pop();
        return res;
    }
    
    int top() {
        return que.back();
    }
    
    bool empty() {
        return que.empty();
    }
};

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