首页 > 技术文章 > LeetCode OJ 之 Delete Node in a Linked List (删除链表中的结点)

jzssuanfa 2017-07-08 09:02 原文

题目:

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.

思路:

把下一个结点的值替换到要删除的结点。然后删除下一个结点。

代码:

/**
 * 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 == NULL || node->next == NULL)
            return;
        ListNode *tmp = node->next;
        node->val = tmp->val;
        node->next = tmp->next;
        delete tmp;
    }
};


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