[Leetcode] 641. Design Circular Deque
2021. 1. 19. 09:52ㆍProblem Solving/LeetCode
Leetcode 641. Design Circular Deque
https://leetcode.com/problems/design-circular-deque/
Design Circular Deque - LeetCode
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Solution
class MyCircularDeque: def __init__(self, k: int): self.q = [None] * (k+1) self.max = k+1 self.front = 0 self.rear = 0 def insertFront(self, value: int) -> bool: if self.isFull(): return False else: self.q[self.front] = value self.front = (self.front - 1 + self.max) % self.max return True def insertLast(self, value: int) -> bool: if self.isFull(): return False else: self.rear = (self.rear + 1) % self.max self.q[self.rear] = value return True def deleteFront(self) -> bool: if self.isEmpty(): return False else: self.q[(self.front + 1) % self.max] = None self.front = (self.front + 1) % self.max return True def deleteLast(self) -> bool: if self.isEmpty(): return False else: self.q[self.rear] = None self.rear = (self.rear - 1 + self.max) % self.max return True def getFront(self) -> int: if self.isEmpty(): return -1 else: return self.q[(self.front + 1) % self.max] def getRear(self) -> int: if self.isEmpty(): return -1 else: return self.q[self.rear] def isEmpty(self) -> bool: return self.front == self.rear def isFull(self) -> bool: return self.front == (self.rear + 1) % self.max # Your MyCircularDeque object will be instantiated and called as such: # obj = MyCircularDeque(k) # param_1 = obj.insertFront(value) # param_2 = obj.insertLast(value) # param_3 = obj.deleteFront() # param_4 = obj.deleteLast() # param_5 = obj.getFront() # param_6 = obj.getRear() # param_7 = obj.isEmpty() # param_8 = obj.isFull()

원형 큐에서 몇가지 코드만 추가하면 된다.
큐는 선입 후출이지만, 덱은 front와 rear모두에서 삽입과 삭제가 가능하다.
주의할 점은, index를 감소시키는 insertFront()
와 deleteLast()
에서
self.front = (self.front - 1 + self.max) % self.max
이렇게 + self.max
를 해줘야 한다는 것이다. self.front
의 인덱스가 0이면 front가 음수가 되기 때문이다.
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