[LeetCode]237. Delete Node in a Linked List ★

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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:
链表
题目大意:删除链表中的指定节点(除了最后一个节点),只给出要删除节点的指针

样例

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.

python解法

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        if not node or not node.next:
            return
        node.val = node.next.val
        node.next = node.next.next

Runtime: 40 ms, faster than 94.25% of Python3 online submissions for Delete Node in a Linked List.
Memory Usage: 13.8 MB, less than 9.38% of Python3 online submissions for Delete Node in a Linked List.
题后反思:无

C语言解法

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

Runtime: 4 ms, faster than 97.59% of C online submissions for Delete Node in a Linked List.
Memory Usage: 7.7 MB, less than 74.07% of C online submissions for Delete Node in a Linked List.
题后反思:

  1. 以前删除节点的方式有点思维定式了,总觉得删除一个节点的时候,指针要指向要删除节点的上一个节点,没有相当过还有这种直接值覆盖的方式。
  2. 但是这种方式有个缺点,就是无法删除最后一个节点

文中都是我个人的理解,如有错误的地方欢迎下方评论告诉我,我及时更正,大家共同进步

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转载自blog.csdn.net/xingyu97/article/details/100066223
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