題目：原題鏈接（中等）
標籤：單鏈表、設計
| 解法 | 時間複雜度 | 空間複雜度 | 執行用時 |
|---|---|---|---|
| Ans 1 (Python) | – | – | 208ms (77.36%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
LeetCode的Python執行用時隨緣，只要時間複雜度沒有明顯差異，執行用時一般都在同一個量級，僅作參考意義。
解法一：
class MyLinkedList:
class _Node:
""""""
__slots__ = "value", "next"
def __init__(self, value, next=None):
self.value = value
self.next = next
def __str__(self):
return str(self.value) + "->" + str(self.next)
def __init__(self):
self._head = self._Node(0)
self._size = 0
def get(self, index: int) -> int:
if index < 0 or index >= self._size:
return -1
curr = self._head
for _ in range(index + 1):
curr = curr.next
return curr.value
def addAtHead(self, val: int) -> None:
self.addAtIndex(0, val)
def addAtTail(self, val: int) -> None:
self.addAtIndex(self._size, val)
def addAtIndex(self, index: int, val: int) -> None:
if index < 0 or index > self._size:
return
self._size += 1
prev = self._head
for _ in range(index):
prev = prev.next
node = self._Node(val, prev.next)
prev.next = node
def deleteAtIndex(self, index: int) -> None:
if index < 0 or index >= self._size:
return
self._size -= 1
prev = self._head
for _ in range(index):
prev = prev.next
prev.next = prev.next.next