[LeetCode] Implement Queue using Stacks 用棧來實現隊列

 

Implement the following operations of a queue using stacks.

• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is empty.

Notes:

• You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
• You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

 

這道題讓我們用棧來實現隊列，之前我們做過一道相反的題目Implement Stack using Queues 用隊列來實現棧，是用隊列來實現棧。這道題顛倒了個順序，起始並沒有太大的區別，棧和隊列的核心不同點就是棧是先進后出，而隊列是先進先出，那么我們要用棧的先進后出的特性來模擬出隊列的先進先出。那么怎么做呢，其實很簡單，只要我們在插入元素的時候每次都都從前面插入即可，比如如果一個隊列是1,2,3,4，那么我們在棧中保存為4,3,2,1，那么返回棧頂元素1，也就是隊列的首元素，則問題迎刃而解。所以此題的難度是push函數，我們需要一個輔助棧tmp，把s的元素也逆着順序存入tmp中，此時加入新元素x，再把tmp中的元素存回來，這樣就是我們要的順序了，其他三個操作也就直接調用棧的操作即可，參見代碼如下：

 

解法一：

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 (!st.empty()) {
            tmp.push(st.top()); st.pop();
        }
        st.push(x);
        while (!tmp.empty()) {
            st.push(tmp.top()); tmp.pop();
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    int pop() {
        int val = st.top(); st.pop();
        return val;
    }
    
    /** Get the front element. */
    int peek() {
        return st.top();
    }
    
    /** Returns whether the queue is empty. */
    bool empty() {
        return st.empty();
    }
    
private:
    stack<int> st;
};

 

上面那個解法雖然簡單，但是效率不高，因為每次在push的時候，都要翻轉兩邊棧，下面這個方法使用了兩個棧_new和_old，其中新進棧的都先緩存在_new中，入股要pop和peek的時候，才將_new中所有元素移到_old中操作，提高了效率，代碼如下：

 

解法二：

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

 

類似題目：

Implement Stack using Queues

 

參考資料：

https://leetcode.com/problems/implement-queue-using-stacks/

https://leetcode.com/problems/implement-queue-using-stacks/discuss/64197/Easy-Java-solution-just-edit-push()-method

https://leetcode.com/problems/implement-queue-using-stacks/discuss/64206/Short-O(1)-amortized-C%2B%2B-Java-Ruby

 

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