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]仅仅知道当前节点。要删除当前节点。和通常的删除不同的是,我们不知道前一个节点。
我们能够利用当前节点与前一个节点的联系,把后一个节点拷贝过来。
然后删除后一个节点,相当于实现了删除操作。
/** * 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) { if(node == nullptr) return; ListNode *next = node->next; node->val = node->next->val; node->next = next->next; delete next; } };