【LeetCode】面试题18. 删除链表的节点

题目:

思路:

单链表的删除(无重复元素)，特殊情况链表为空或链表头元素为要删除的元素(此时返回的head修改了)，其它情况记录上个节点和当前节点，删除后返回head即可。

代码:

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def deleteNode(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        if head is None:
            return None
        if head.val == val:
            return head.next
        preElem = head
        nowElem = head.next
        while nowElem is not None:
            if nowElem.val == val:
                preElem.next = nowElem.next
                nowElem.next = None
                return head
            else:
                preElem = nowElem
                nowElem = nowElem.next
        return head

相关问题

【LeetCode】面试题18. 删除链表的节点

原文：https://www.cnblogs.com/cling-cling/p/12986009.html