剑指 Offer 09. 用两个栈实现队列

Posted 2022-04-28 Billy Miracle

题目描述：

思路：

这玩意不难，但容易想复杂，比如我，我先是保证每次添加后输出栈都是按照队列的顺序了，下面一会儿会给出代码，显然这样会比较麻烦，多遍历了。标准答案就简单了，只在输出栈为空后把输入栈重新压入就好了。

代码：

//傻了吧唧版：
class CQueue 
private:
    stack<int> inStack, outStack;

    void inToOut() 
        while (!inStack.empty()) 
            outStack.push(inStack.top());
            inStack.pop();
        
    
    void outToIn() 
        while (!outStack.empty()) 
            inStack.push(outStack.top());
            outStack.pop();
        
    

public:
    CQueue() 

    
    
    void appendTail(int value) 
        outToIn();
        inStack.push(value);
        inToOut();
    
    
    int deleteHead() 
        if (outStack.empty()) 
            return -1;
        
        int value = outStack.top();
        outStack.pop();
        return value;
    
;

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

//标准答案：
class CQueue 
private:
    stack<int> inStack, outStack;

    void in2out() 
        while (!inStack.empty()) 
            outStack.push(inStack.top());
            inStack.pop();
        
    

public:
    CQueue() 

    void appendTail(int value) 
        inStack.push(value);
    

    int deleteHead() 
        if (outStack.empty()) 
            if (inStack.empty()) 
                return -1;
            
            in2out();
        
        int value = outStack.top();
        outStack.pop();
        return value;
    
;
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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