[LeetCode] 237. Delete Node in a Linked List 删除链表的节点

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:

    4 -> 5 -> 1 -> 9

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.

写一个函数删除单链表中的一个节点，链表至少有2个元素，所有的节点值是唯一的，给的节点不会是尾部并且是合法的节点，不返回任何值。要求in-place。

解法：先把next节点的值赋给当前节点，在把当前节点的next变成当前节点的next.next。

Java：

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public class Solution {     public void deleteNode(ListNode node) {         node.val = node.next.val;         node.next = node.next.next;     } }

Python:

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def deleteNode(self, node):         """         :type node: ListNode         :rtype: void Do not return anything, modify node in-place instead.         """         node.val = node.next.val         node.next = node.next.next

Python：

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class Solution:     # @param {ListNode} node     # @return {void} Do not return anything, modify node in-place instead.     def deleteNode(self, node):         if node and node.next:             node_to_delete = node.next             node.val = node_to_delete.val             node.next = node_to_delete.next             del node_to_delete

C++:

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void deleteNode(ListNode* node) {     *node = *node->next; }

C++：

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class Solution { public:     void deleteNode(ListNode* node) {         node->val = node->next->val;         ListNode *tmp = node->next;         node->next = tmp->next;         delete tmp;     } };

JavaScript:

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var deleteNode = function(node) {     node.val = node.next.val;     node.next = node.next.next; };

Ruby:　　

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def delete_node(node)     node.val = node.next.val     node.next = node.next.next     nil end

　　

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[LeetCode] 203. Remove Linked List Elements 移除链表元素

　

　

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