leetcode[237]:Delete Node in a Linked List

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本文介绍了一种在单链表中删除指定节点的方法，当无法访问前驱节点时，可以通过覆盖当前节点值并跳过其后继节点来实现删除操作。

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Delete Node in a Linked List

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.

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

找不到前驱，可以删除其后继，其后继的值赋予当前结点即可。