237 Delete Node in a Linked List 删除链表的结点

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;
    }
};

  

原文地址:https://www.cnblogs.com/xidian2014/p/8759426.html