[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：

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

Python:

# 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：

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++:

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

C++：

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

JavaScript:

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

Ruby:　　

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

　　

类似题目：

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

　

　

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