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  • LeetCode OJ 之 Delete Node in a Linked List (删除链表中的结点)

    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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  • 原文地址:https://www.cnblogs.com/jzssuanfa/p/7135817.html
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