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.
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解法：题目要求我们删除链表中的一个节点，但是并不告诉我们链表的头节点，而只给我们那个要删除的节点。因为这个节点不是尾节点，因此可以先将待删除节点的下一个节点的值保存到此节点，然后这个下一个节点即可。这也是单链表操作中一个常用的技巧。
/** * 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) { ListNode* del = node->next; node->val = node->next->val; node->next = node->next->next; delete del; } };
[LeetCode]79. Delete Node in a Linked List删除链表节点
原文：http://www.cnblogs.com/aprilcheny/p/4964895.html