题解 | #从单向链表中删除指定值的节点#

# 数据结构
class ListNode:
    def __init__(self, data=None， next=None):
        self.data = data
        self.next = next

# 找到某个节点        
def findVal(head: ListNode, value: str) -> ListNode:
    while head:
        if head.data == value:
            return head
        else:
            head = head.next
    return None

# 逐个打印所有节点
def printAll(head: ListNode) -> None:
    while head:
        print(head.data, end=" ")
        head = head.next

# 删除一个节点
def remove(head: ListNode, value: int) -> bool:
    while head:
        if head.next.data == value:
            head.next = head.next.next
            return True
        else:
            head = head.next
    return False


s = input()
s = list(map(int, s.strip().split()))
listNodeLength = s[0] # 第一个数字是长度
firtsListNodeData = s[1] # 第二个数字是第一个Node的值
appndOp = s[2:listNodeLength * 2:] # 中间值为要掺入的操作集
appendant = appndOp[::2] # 要被加入的Node值
base = appndOp[1::2] # 在这些Node后面加值
head = ListNode(None) # dummy Node，时刻防止丢失头节点
head.next = ListNode(firtsListNodeData) # 第一个节点

# 增
for i in range(len(base)):
    temp = findVal(head, base[i])
    oriNext = temp.next
    temp.next = ListNode(appendant[i])
    temp.next.next = oriNext

remove(head, s[-1]) # 删
printAll(head.next) # 打印