LeetCode每日一题-5：删除排序链表中的重复元素

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        # 快慢指针
        if not head:
            return None
        fast, slow = head, head
        while fast:
            if fast.val != slow.val:
                slow.next = fast
                slow = slow.next
            fast = fast.next
        slow.next = None
        return head

public ListNode deleteDuplicates(ListNode head) {
    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null){
            return head;
        }
        ListNode dummhead=new ListNode(0);
        dummhead.next=head;
        ListNode cur=head;
        ListNode last=head.next;
        while(last!=null){
            if(cur.val==last.val){
                ListNode tmp=null;
                tmp=last.next;
                cur.next=tmp;
                last=tmp;
            }else{
                cur=last;
                last=last.next;
            }
        }
        return dummhead.next;
    }
}