一、题目描述

题目描述

二、思路

声明两个栈解决：
stack<int> st1用于存储；
stack<int> st2用于辅助操作。
push()实现：直接存入st1；
pop()和peak()实现：先把st1元素依次存入st2，然后获取st2的栈顶元素，操作之后将st2的元素又存入st1。

三、代码

class MyQueue {
public:
    stack<int> st1;    // 用于存储
    stack<int> st2;    // 用于辅助 
    
    /** Initialize your data structure here. */
    MyQueue() {
        
    }
    
    /** Push element x to the back of queue. */
    void push(int x) {
        st1.push(x);
        
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        while(st1.size() != 0){
            int cur = st1.top();
            st2.push(cur);
            st1.pop();
        }
        int res = st2.top();
        st2.pop();
        while(st2.size() != 0){
            int cur = st2.top();
            st1.push(cur);
            st2.pop();
        }
        return res;
    }
    
    /** Get the front element. */
    int peek() {
        while(st1.size() != 0){
            int cur = st1.top();
            st2.push(cur);
            st1.pop();
        }
        int res = st2.top();
        while(st2.size() != 0){
            int cur = st2.top();
            st1.push(cur);
            st2.pop();
        }
        return res;
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        if(st1.size() == 0)
            return true;
        return false;
    }
};

/**
 * 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();
 */

四、提交结果

提交结果