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