203. 移除链表元素
题意
删除链表中等于给定值 val 的所有节点；
解题思路
把下个结点的值赋值给当前结点，删除下一个结点；
增加一个多余的头结点，方便记录下前结点，将前结点指向下个结点，删除当前结点；
实现
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
node = head
while node and node.val != val:
node = node.next
while node:
if node.val == val:
# 把下个结点的值赋值给当前结点，实际是删除下一个结点
if node.next != None:
node.val = node.next.val
node.next = node.next.next
elif head == node: # 头结点
head = None
break
else: # 尾结点
cur = head
while cur.next and cur.next != node:
cur = cur.next
cur.next = None
break
else:
node = node.next
return head
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
# 新增一个起始结点，方便获取前结点（比如在删除首结点的时候）
start = ListNode(0)
start.next = head
prev = start
cur = head
while cur is not None:
if cur.val != val:
prev = cur
cur = cur.next
else:
prev.next = cur.next
cur = prev.next
return start.next
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
try:
while head.val == val:
head = head.next
cur = head
nex = cur.next
while nex:
if nex.val != val:
cur.next = nex
cur = cur.next
nex = nex.next
if cur.next.val == val:
cur.next = None
except:
True
return head