``` 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. 赋值法 复杂度 时间 O(1) 空间 O(1) 思路 乍一看没法获取上一个链表节点，似乎无法将当前结点去除。实际上只要将下一个节点的值覆盖当前节点，然后删除下一个节点就好了。注意这样不适用于尾节点。 代码 public class Solution { public void deleteNode(ListNode node) { node.val = node.next.val; node.next = node.next.next; } } Remove Linked List Elements 伪造表头 复杂度 时间 O(N) 空间 O(1) 思路 删除链表所有的特定元素的难点在于如何处理链表头，如果给加一个dummy表头，然后再从dummy表头开始遍历，最后返回dummy表头的next，就没有这么问题了。 代码 public class Solution { public ListNode removeElements(ListNode head, int val) { ListNode dummy = new ListNode(-1); dummy.next = head; ListNode curr = dummy; while(curr.next!=null){ if(curr.next.val == val){ curr.next = curr.next.next; } else { curr = curr.next; } } return dummy.next; } } ```

# [Leetcode] Delete Node/Remove Element in a Linked List 删除链表节点

http://www.js-code.com/node-js/node-js_35612.html