编写一个函数,在给定单链表一个结点(非尾结点)的情况下,删除该结点。

假设该链表为1 -> 2 -> 3 -> 4 并且给定你链表中第三个值为3的节点,在调用你的函数后,该链表应变为1 -> 2 -> 4。

详见:https://leetcode.com/problems/delete-node-in-a-linked-list/description/

Java实现:

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void deleteNode(ListNode node) {
node.val=node.next.val;
node.next=node.next.next;
}
}

C++实现:

/**
* 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 *tmp=node->next;
node->next=tmp->next;
delete tmp;
}
};

  

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