237. Delete Node in a Linked List（python+cpp）

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:
4 -> 5 -> 1 -> 9
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.

解释：
删除单链表中给定的结点。一般删除单链表中的一个结点都是给定它的前一个结点，但是这里是给定指定结点，那么怎么删除呢？方法就是不删除这个结点，而是吧它的后一个结点的值给它，然后删除它的后一个结点，从结果上来看就是删除了当前结点，而且时间复杂度是O(1)（注意，用这种方法时，所给的结点不能是最后一个结点，注意题目的Note）。

python代码：

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

class Solution(object):
    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

c++代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        node->val=node->next->val;
        node->next=node->next->next;
    }
};

总结：
用O(1)的时间复杂度删除指定结点。