题目:
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 -> 4after calling your function.
思路:
题目只给出了要删除的节点,没有给出前指针或根节点,因此只能利用该节点和其之后的节点实现删除操作。
1.把该节点后一个的节点的值赋给该节点。假如要删除的节点为3,经过赋值操作后为:1->2->4->4
2.删除该节点后一个节点,原节点为:1->2->4
代码:
1 class Solution {
2 public:
3 void deleteNode(ListNode*
node) {
4 if (!node || !node->
next) {
5 return;
6 }
7 ListNode *nextNode = node->
next;
8 node->val = node->next->
val;
9 node->next = node->next->
next;
10 }
11 };
转载于:https://www.cnblogs.com/sindy/p/6805802.html