用Python實現雙向連結串列
阿新 • • 發佈:2018-11-11
直接看程式碼,有註解
class Node(object): """結點""" def __init__(self, item): self.elem = item self.next = None self.prev = None class DoubleLinkList(object): """雙鏈表""" def __init__(self, node=None): self.__head = node def is_empty(self): """連結串列是否為空""" return self.__head == None def length(self): """連結串列長度""" # cur遊標,用來移動遍歷節點 cur = self.__head # count記錄數量 count = 0 while cur != None: count += 1 cur = cur.next return count def travel(self): """遍歷整個連結串列""" cur = self.__head while cur != None: print(cur.elem, end=" ") cur = cur.next print("") def add(self, item): """連結串列頭部新增元素,頭插法""" node = Node(item) node.next = self.__head self.__head = node node.next.prev = node def append(self, item): """連結串列尾部新增元素, 尾插法""" node = Node(item) if self.is_empty(): self.__head = node else: cur = self.__head while cur.next != None: cur = cur.next cur.next = node node.prev = cur def insert(self, pos, item): """指定位置新增元素 :param pos 從0開始 """ if pos <= 0: self.add(item) elif pos > (self.length()-1): self.append(item) else: cur = self.__head count = 0 while count < pos: count += 1 cur = cur.next # 當迴圈退出後,cur指向pos位置 node = Node(item) node.next = cur node.prev = cur.prev cur.prev.next = node cur.prev = node def remove(self, item): """刪除節點""" cur = self.__head while cur != None: if cur.elem == item: # 先判斷此結點是否是頭節點 # 頭節點 if cur == self.__head: self.__head = cur.next if cur.next: # 判斷連結串列是否只有一個結點 cur.next.prev = None else: cur.prev.next = cur.next if cur.next: cur.next.prev = cur.prev break else: cur = cur.next def search(self, item): """查詢節點是否存在""" cur = self.__head while cur != None: if cur.elem == item: return True else: cur = cur.next return False if __name__ == "__main__": ll = DoubleLinkList() print(ll.is_empty()) print(ll.length()) ll.append(1) print(ll.is_empty()) print(ll.length()) ll.append(2) ll.add(8) ll.append(3) ll.append(4) ll.append(5) ll.append(6) # 8 1 2 3 4 5 6