LeetCode#237 Delete Node in a Linked List

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Problem Definition:

　　Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

　　Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

思路：

　　本题的难点在于，要在单链表中删除某个节点，且只有对这个要删除的节点的访问。由于无法访问这样要删除节点前面的节点，常规的节点删除方法不适用。

　　怎么利用对当前要删除的节点以及它的后续节点来完成删除操作嘞？

　　把后续节点的值复制到要删除的节点，再把这个后续节点删除。

代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} node
    # @return {void} Do not return anything, modify node in-place instead.
    def deleteNode(self, node):
        if node.next==None:
            return
        node.val=node.next.val
        node.next=node.next.next
        

注意边界情况的处理：如果要删除的是最后一个节点，直接跳过。

LeetCode#237 Delete Node in a Linked List

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原文地址：http://www.cnblogs.com/acetseng/p/4648657.html