Leetcode:Delete Node in a Linked List
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.Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.
Answer
目標是刪除當前給你的 Node ,於是乎只要把下一個的 Node 資訊全部 Copy & Paste 到當前 Node 即可。Javascript
_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_//**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} node
* @return {void} Do not return anything, modify node in-place instead.
*/
var deleteNode = function(node){
node.val = node.next.val;
node.next = node.next.next;
}
//Min:132ms
_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/C
_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_//**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node){
node->val = node->next->val;
node->next = node->next->next;
}
//Min:4ms
_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/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
#Min:48ms
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