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;
    }
}

本文固定链接: http://www.js-code.com/node-js/node-js_33157.html