641. Design Circular Deque

 

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.

 

这道题让我们设计一个环形双向队列,由于之前刚做过一道Design Circular Queue,那道设计一个环形队列,其实跟这道题非常的类似,环形双向队列在环形队列的基础上多了几个函数而已,其实本质并没有啥区别,那么之前那道题的解法一改吧改吧也能用在这道题上,参见代码如下:

 

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;
};

 

就像前一道题中的分析的一样,上面的解法并不是本题真正想要考察的内容,我们要用上环形Circular的性质,我们除了使用size来记录环形队列的最大长度之外,还要使用三个变量,head,tail,cnt,分别来记录队首位置,队尾位置,和当前队列中数字的个数,这里我们将head初始化为k-1,tail初始化为0。还是从简单的做起,判空就看当前个数cnt是否为0,判满就看当前个数cnt是否等于size。接下来取首尾元素,先进行判空,然后根据head和tail分别向后和向前移动一位取即可,记得使用上循环数组的性质,要对size取余。再来看删除末尾函数,先进行判空,然后tail向前移动一位,使用循环数组的操作,然后cnt自减1。同理,删除开头函数,先进行判空,队首位置head要向后移动一位,同样进行加1之后对长度取余的操作,然后cnt自减1。再来看插入末尾函数,先进行判满,然后将新的数字加到当前的tail位置,tail移动到下一位,为了避免越界,我们使用环形数组的经典操作,加1之后对长度取余,然后cnt自增1即可。同样,插入开头函数,先进行判满,然后将新的数字加到当前的head位置,head移动到前一位,然后cnt自增1,参见代码如下:

 

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;
};

 

论坛上还见到了使用链表来做的解法,由于博主比较抵触在解法中新建class,所以这里就不贴了,可以参见这个帖子

 

类似题目:

Design Circular Queue

 

参考资料:

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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