LeetCode 237. Delete Node in a Linked List
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描述
删除链表中的一个节点
解决
参数是指向链表的一个指针。如果node = node -> next
,改变的只是该指针的值,而没有实际改变链表中的节点, *node = *(node -> next)
, 才是正确的。
/** * 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 = *(node -> next); }};
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