Delete Node in a Linked List
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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.
这道题让我们删除链表的一个节点,更通常不同的是,没有给我们链表的起点,只给我们了一个要删的节点,跟我们以前遇到的情况不太一样,我们之前要删除一个节点的方法是要有其前一个节点的位置,然后将其前一个节点的next连向要删节点的下一个,然后delete掉要删的节点即可。这道题的处理方法是先把当前节点的值用下一个节点的值覆盖了,然后我们删除下一个节点即可:
/** * 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; ListNode* temp=node->next; node->next=temp->next; free(temp); }};
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