[LeetCode] Design Circular Deque 设计环形双向队列

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.
 

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 -;
return data.front();
} /** Get the last item from the deque. */
int getRear() {
if (isEmpty()) return -;
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 - ; tail = , cnt = ;
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 - + 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 + ) % 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 + ) % 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 - + size) % size;
--cnt;
return true;
} /** Get the front item from the deque. */
int getFront() {
return isEmpty() ? - : data[(head + ) % size];
} /** Get the last item from the deque. */
int getRear() {
return isEmpty() ? - : data[(tail - + size) % size];
} /** Checks whether the circular deque is empty or not. */
bool isEmpty() {
return cnt == ;
} /** 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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