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[LeetCode] 232. 用栈实现队列 #45

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Animenzzzz opened this issue Aug 14, 2019 · 0 comments
Open

[LeetCode] 232. 用栈实现队列 #45

Animenzzzz opened this issue Aug 14, 2019 · 0 comments

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@Animenzzzz
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题目描述:

使用栈实现队列的下列操作:

push(x) -- 将一个元素放入队列的尾部。
pop() -- 从队列首部移除元素。
peek() -- 返回队列首部的元素。
empty() -- 返回队列是否为空。

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

说明:

你只能使用标准的栈操作 -- 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
你所使用的语言也许不支持栈。你可以使用 list 或者 deque(双端队列)来模拟一个栈,只要是标准的栈操作即可。
假设所有操作都是有效的 (例如,一个空的队列不会调用 pop 或者 peek 操作)。

解题思路:使用一个中间导数的栈,这个tmp栈用于暂时存放数据,当push时,先把原栈导出清空,把push的元素入栈,然后再把tmp栈放回

C++解题:

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