[Leetcode]203. Remove Linked List Elements

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

 

思路：我们看当前节点的下一个是否等于val，如果等于的话，就把当前节点的next指向下个节点的next

/**
 * 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) {
        ListNode current = head;
        while (current!=null){
            if (current==head&&current.val==val){
                head = head.next;     //如果是头节点的话，就移动头节点
                current = head;
            }
            else if (current.next!=null&&current.next.val==val)
                current.next = current.next.next;  //看当前节点的下一个的值是否为val
            else
                current = current.next;
        }
    }
}