leetcode --237 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.
题目简单。思想不简单。
常规办法是从头检索到结点,然后将前一节点的next指针指向删除结点的后一个节点;但是这道题没有给头指针,因此不能这么干。
由于不是尾结点,可以将待删除结点的后一个节点的值复制到当前结点,然后将当前结点的next指针指向后一节点的next即可。
时间15ms。
代码如下:
/** * 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)//判空 return; ListNode* nextnode=node->next; node->val = nextnode->val; node->next = nextnode->next; }};
可以不申请新的指针,直接使用原来的指针。
class Solution {public: void deleteNode(ListNode* node) { if(!node) return; //ListNode* nextnode=node->next; node->val = node->next->val; node->next = node->next->next; }};
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