237 - Delete Node in a Linked List

删除链表节点

最新推荐文章于 2019-03-01 22:03:47 发布

原创最新推荐文章于 2019-03-01 22:03:47 发布 · 518 阅读

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本文介绍了一个函数，该函数用于删除单链表中的指定节点（除尾节点外），仅需访问该节点。通过修改节点值及指向下一个节点的指针实现删除操作。

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.

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void deleteNode(struct ListNode* node) {
    if(node != NULL && node->next != NULL) {
        node->val = node->next->val;
        node->next = node->next->next;
    }
    
}