剑指 Offer 09. 用两个栈实现队列

剑指 Offer 09. 用两个栈实现队列

难度简单249

用两个栈实现一个队列。队列的声明如下，请实现它的两个函数 appendTail 和 deleteHead ，分别完成在队列尾部插入整数和在队列头部删除整数的功能。(若队列中没有元素，deleteHead 操作返回 -1 )

示例 1：

输入：
["CQueue","appendTail","deleteHead","deleteHead"]
[[],[3],[],[]]
输出：[null,null,3,-1]

示例 2：

输入：
["CQueue","deleteHead","appendTail","appendTail","deleteHead","deleteHead"]
[[],[],[5],[2],[],[]]
输出：[null,-1,null,null,5,2]

提示：

• 1 <= values <= 10000
• 最多会对 appendTail、deleteHead 进行 10000 次调用

法一：欣赏一下我写的垃圾玩意

class CQueue {
public:
    CQueue() {

    }
    
    void appendTail(int value) {
        s1.push(value);
        if(s2.empty()){
            while(!s1.empty()){
                s2.push(s1.top());
                s1.pop();
            }
        }
    }
    
    int deleteHead() {
        int val=-1;
        if(!s2.empty()){
            val =  s2.top();
            s2.pop();
        }
        if(s2.empty())
            while(!s1.empty()){
                s2.push(s1.top());
                s1.pop();
            }
        
        return val;
    }
private:
    stack<int>s1, s2;
};

/**
 * Your CQueue object will be instantiated and called as such:
 * CQueue* obj = new CQueue();
 * obj->appendTail(value);
 * int param_2 = obj->deleteHead();
 */

法二：优雅的官解

class CQueue {
    stack<int> stack1,stack2;
public:
    CQueue() {
        while (!stack1.empty()) {
            stack1.pop();
        }
        while (!stack2.empty()) {
            stack2.pop();
        }
    }
    
    void appendTail(int value) {
        stack1.push(value);
    }
    
    int deleteHead() {
        // 如果第二个栈为空
        if (stack2.empty()) {
            while (!stack1.empty()) {
                stack2.push(stack1.top());
                stack1.pop();
            }
        } 
        if (stack2.empty()) {
            return -1;
        } else {
            int deleteItem = stack2.top();
            stack2.pop();
            return deleteItem;
        }
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/yong-liang-ge-zhan-shi-xian-dui-lie-lcof/solution/mian-shi-ti-09-yong-liang-ge-zhan-shi-xian-dui-l-3/
来源：力扣（LeetCode）
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