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

時間 2019-11-16

標籤 leetcode implement queue using stacks 實現隊列简体版

原文原文鏈接

Implement the following operations of a queue using stacks.html

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:緩存

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).

這道題讓咱們用棧來實現隊列，以前咱們作過一道相反的題目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，起始並無太大的區別，棧和隊列的核心不一樣點就是棧是先進後出，而隊列是先進先出，那麼咱們要用棧的先進後出的特性來模擬出隊列的先進先出。那麼怎麼作呢，其實很簡單，只要咱們在插入元素的時候每次都都從前面插入便可，好比若是一個隊列是1,2,3,4，那麼咱們在棧中保存爲4,3,2,1，那麼返回棧頂元素1，也就是隊列的首元素，則問題迎刃而解。因此此題的難度是push函數，咱們須要一個輔助棧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，這樣就是咱們要的順序了，其餘三個操做也就直接調用棧的操做便可，參見代碼以下：函數

解法一：post

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，可是效率不高，由於每次在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，纔將_new中全部元素移到_old中操做，提升了效率，代碼以下：url

解法二：spa

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;
};