LeetCode #237 - Delete Node in a Linked List

最新推荐文章于 2021-03-15 23:29:59 发布

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本文介绍了一种删除链表中指定节点的方法，不返回任何值。示例展示了如何将链表4->5->1->9变为4->1->9或4->5->9，通过遍历并更新节点值实现。

题目描述：

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following: 4 -> 5 -> 1 -> 9

Example 1:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
             should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list
             should become 4 -> 5 -> 9 after calling your function.

Note:

The linked list will have at least two elements.
All of the nodes' values will be unique.
The given node will not be the tail and it will always be a valid node of the linked list.
Do not return anything from your function.

移除链表中的一个节点，直接遍历查找并删除。

class Solution {
public:
    void deleteNode(ListNode* node) {
        ListNode* p=node;
        ListNode* pre_p=new ListNode(0);
        while(p->next!=NULL)
        {
            pre_p=p;
            p->val=p->next->val;
            p=p->next;
        }
        pre_p->next=NULL;
    }
};