5-237. Delete Node in a Linked List

题目描述：

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

Given linked list -- head = [4,5,1,9], which looks like following:

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes' values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

代码实现（大神版）：

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

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val
        node.next = node.next.next

分析：

关于用Python实现链表的知识详见：https://www.cnblogs.com/kumata/p/9147077.html

题目的要求是给定一个节点（该节点不在链表的尾部），将该节点删除掉。

实现思路非常简单：只需要用下一个节点来覆盖要删除的节点即可，也就是让当前节点的内容等于下一个节点的内容，同时让当前节点的指针指向下一个节点的指针即可。

如果给出的不是一个节点，而是给出节点中的内容，让删除这个节点，则一种代码实现方式如下：（个人版）

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        a = ListNode(9)
        b = ListNode(1)
        c = ListNode(5)
        d = ListNode(4)

        b.next = a
        c.next = b
        d.next = c
        head = d

        fpointer, bpointer = head, head
        while True:
            if fpointer.val == node:
                if fpointer == head:
                    head.val = fpointer.next.val
                    head.next = fpointer.next.next
                elif fpointer.next == None:
                    bpointer.next = None
                else:
                    bpointer.next = fpointer.next
                    bpointer.next.val = fpointer.next.val
                break
            else:
                bpointer = fpointer
                fpointer = fpointer.next
        
        pointer = head
        aa = [0] * 3
        i = 0
        while i < 3:
            aa[i] = pointer.val
            pointer = pointer.next
            i += 1
        print(aa)

if __name__ == '__main__':
    Solution().deleteNode(9)