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。
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参考资料：