[LeetCode] Design Circular Deque 設計環形雙向隊列

 

Design your implementation of the circular double-ended queue (deque).

Your implementation should support following operations:

• MyCircularDeque(k): Constructor, set the size of the deque to be k.
• insertFront(): Adds an item at the front of Deque. Return true if the operation is successful.
• insertLast(): Adds an item at the rear of Deque. Return true if the operation is successful.
• deleteFront(): Deletes an item from the front of Deque. Return true if the operation is successful.
• deleteLast(): Deletes an item from the rear of Deque. Return true if the operation is successful.
• getFront(): Gets the front item from the Deque. If the deque is empty, return -1.
• getRear(): Gets the last item from Deque. If the deque is empty, return -1.
• isEmpty(): Checks whether Deque is empty or not. 
• isFull(): Checks whether Deque is full or not.

 

Example:

MyCircularDeque circularDeque = new MycircularDeque(3); // set the size to be 3
circularDeque.insertLast(1);			// return true
circularDeque.insertLast(2);			// return true
circularDeque.insertFront(3);			// return true
circularDeque.insertFront(4);			// return false, the queue is full
circularDeque.getRear();  			// return 2
circularDeque.isFull();				// return true
circularDeque.deleteLast();			// return true
circularDeque.insertFront(4);			// return true
circularDeque.getFront();			// return 4

 

Note:

• All values will be in the range of [0, 1000].
• The number of operations will be in the range of [1, 1000].
• Please do not use the built-in Deque library.

 

這道題讓我們設計一個環形雙向隊列，由於之前剛做過一道Design Circular Queue，那道設計一個環形隊列，其實跟這道題非常的類似，環形雙向隊列在環形隊列的基礎上多了幾個函數而已，其實本質並沒有啥區別，那么之前那道題的解法一改吧改吧也能用在這道題上，參見代碼如下：

 

解法一：

class MyCircularDeque {
public:
    /** Initialize your data structure here. Set the size of the deque to be k. */
    MyCircularDeque(int k) {
        size = k;   
    }
    
    /** Adds an item at the front of Deque. Return true if the operation is successful. */
    bool insertFront(int value) {
        if (isFull()) return false;
        data.insert(data.begin(), value);
        return true;
    }
    
    /** Adds an item at the rear of Deque. Return true if the operation is successful. */
    bool insertLast(int value) {
        if (isFull()) return false;
        data.push_back(value);
        return true;
    }
    
    /** Deletes an item from the front of Deque. Return true if the operation is successful. */
    bool deleteFront() {
        if (isEmpty()) return false;
        data.erase(data.begin());
        return true;
    }
    
    /** Deletes an item from the rear of Deque. Return true if the operation is successful. */
    bool deleteLast() {
        if (isEmpty()) return false;
        data.pop_back();
        return true;
    }
    
    /** Get the front item from the deque. */
    int getFront() {
        if (isEmpty()) return -1;
        return data.front();
    }
    
    /** Get the last item from the deque. */
    int getRear() {
        if (isEmpty()) return -1;
        return data.back();
    }
    
    /** Checks whether the circular deque is empty or not. */
    bool isEmpty() {
        return data.empty();
    }
    
    /** Checks whether the circular deque is full or not. */
    bool isFull() {
        return data.size() >= size;
    }

private:
    vector<int> data;
    int size;
};

 

就像前一道題中的分析的一樣，上面的解法並不是本題真正想要考察的內容，我們要用上環形Circular的性質，我們除了使用size來記錄環形隊列的最大長度之外，還要使用三個變量，head，tail，cnt，分別來記錄隊首位置，隊尾位置，和當前隊列中數字的個數，這里我們將head初始化為k-1，tail初始化為0。還是從簡單的做起，判空就看當前個數cnt是否為0，判滿就看當前個數cnt是否等於size。接下來取首尾元素，先進行判空，然后根據head和tail分別向后和向前移動一位取即可，記得使用上循環數組的性質，要對size取余。再來看刪除末尾函數，先進行判空，然后tail向前移動一位，使用循環數組的操作，然后cnt自減1。同理，刪除開頭函數，先進行判空，隊首位置head要向后移動一位，同樣進行加1之后對長度取余的操作，然后cnt自減1。再來看插入末尾函數，先進行判滿，然后將新的數字加到當前的tail位置，tail移動到下一位，為了避免越界，我們使用環形數組的經典操作，加1之后對長度取余，然后cnt自增1即可。同樣，插入開頭函數，先進行判滿，然后將新的數字加到當前的head位置，head移動到前一位，然后cnt自增1，參見代碼如下：

 

解法二：

class MyCircularDeque {
public:
    /** Initialize your data structure here. Set the size of the deque to be k. */
    MyCircularDeque(int k) {
        size = k; head = k - 1; tail = 0, cnt = 0;
        data.resize(k);
    }
    
    /** Adds an item at the front of Deque. Return true if the operation is successful. */
    bool insertFront(int value) {
        if (isFull()) return false;
        data[head] = value;
        head = (head - 1 + size) % size;
        ++cnt;
        return true;
    }
    
    /** Adds an item at the rear of Deque. Return true if the operation is successful. */
    bool insertLast(int value) {
        if (isFull()) return false;
        data[tail] = value;
        tail = (tail + 1)  % size;
        ++cnt;
        return true;
    }
    
    /** Deletes an item from the front of Deque. Return true if the operation is successful. */
    bool deleteFront() {
        if (isEmpty()) return false;
        head = (head + 1) % size;
        --cnt;
        return true;
    }
    
    /** Deletes an item from the rear of Deque. Return true if the operation is successful. */
    bool deleteLast() {
        if (isEmpty()) return false;
        tail = (tail - 1 + size) % size;
        --cnt;
        return true;
    }
    
    /** Get the front item from the deque. */
    int getFront() {
        return isEmpty() ? -1 : data[(head + 1) % size];
    }
    
    /** Get the last item from the deque. */
    int getRear() {
        return isEmpty() ? -1 : data[(tail - 1 + size) % size];
    }
    
    /** Checks whether the circular deque is empty or not. */
    bool isEmpty() {
        return cnt == 0;
    }
    
    /** Checks whether the circular deque is full or not. */
    bool isFull() {
        return cnt == size;
    }

private:
    vector<int> data;
    int size, head, tail, cnt;
};

 

論壇上還見到了使用鏈表來做的解法，由於博主比較抵觸在解法中新建class，所以這里就不貼了，可以參見這個帖子。

 

類似題目：

Design Circular Queue

 

參考資料：

https://leetcode.com/problems/design-circular-deque/

https://leetcode.com/problems/design-circular-deque/discuss/149371/Java-doubly-LinkedList-solution-very-straightforward

https://leetcode.com/problems/design-circular-deque/discuss/155209/c%2B%2B-99-ring-buffer-no-edge-cases.-fb-interviewer-really-loves-it.-easy-to-impl-in-4mins.-cheers!

 

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