LeetCode算法日记：面试题 03.04. 化栈为队

面试题 03.04. 化栈为队

日期：2022/8/2

题目描述：实现一个MyQueue类，该类用两个栈来实现一个队列。

示例：

MyQueue queue = new MyQueue();

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

思路：

peek或pop的时候，用s2存储s1的逆序，这样s2的尾就是s1的首，也就是需要输出的那个元素

代码+解析：

class MyQueue {
private:
    stack<int> s1;
public:
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        s1.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        stack<int> temp = s1;
        stack<int> s2;
        while(!s1.empty()){
            s2.push(s1.top());
            s1.pop();
        }
        int res = s2.top();
        s2.pop();
        while(s1.size() < temp.size()-1){
            s1.push(s2.top());
            s2.pop();
        }
        return res;
    }
    
    /** Get the front element. */
    int peek() {
        stack<int> temp = s1;
        stack<int> s2;
        while(!temp.empty()){
            s2.push(temp.top());
            temp.pop();
        }
        return s2.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return s1.size() == 0;
    }
};

/**
 * 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();
 */