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（除尾节点外）的方法，仅通过访问该节点实现。通过将下一节点的值复制到当前节点并调整指针链接，最终释放下一节点完成删除操作。

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

思想：利用覆盖的理念，将下一个节点覆盖当前节点并将当前节点的next链接下一节点的next，同时free原来当前节点的下一个节点

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