LeetCode刷题Easy篇删除单链表中的元素Delete Node in a Linked List

题目

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.

这道题目以前做过一次，写完后发现不一样，这个没有给头节点，要直接删除指定node。

为了复习这个算法，我先写了一下有head节点的情况，代码出现了几个小问题。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if(head==null) return null;
        ListNode dummy=new ListNode(0);
        dummy.next=head;
        ListNode curr=dummy;
        while(curr!=null&&curr.next!=null){
            if(curr.next.val==val){
                curr.next=curr.next.next;
            }
            else{
                //不相等才移动curr，相等不用
                curr=curr.next;
            }
            
        }
        return dummy.next;

    }
}

此代码在后来学习单链表遍历，利用dummy node 处理corner case之后修改过，完美通过。