LeetCode 237. Delete Node in a Linked List(删除链表中的结点) -- c语言

237. 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.

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

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.

解题思路：

 

//执行时间12ms，战胜90.64%的提交者
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void deleteNode(struct ListNode* node) {//依题意，node是需要删除的结点，
    
    struct ListNode *p = node->next;
    node->val = p->val;//node的值需删除，p的值是需要保存的，而我们是通过删除p来实现删除node的
    node->next = p->next;//剥离出p要删除结点
    free(p);
}

//执行用时24，战胜9.62%的提交者
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
//数组的思维，未利用链表的特点，不可取
void deleteNode(struct ListNode* node) {
    struct ListNode *p = node;
    struct ListNode *q = node->next;
    while(q != NULL)
    {
        node->val = q->val;
        p = node;
        node = node->next;
        q = node->next;
    }
    //此时q指向尾节点的后继，即NULL；node指向尾节点，p指向尾节点的前驱
    p->next = NULL;//断开尾节点，即删除
}

后记：

如果不考虑free释放结点，则可简单成两条语句：node->val = node->next->val;node ->next = node->next->next;