head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
Input: head = [1,2,3,3,4,4,5] Output: [1,2,5]
Input: head = [1,1,1,2,3] Output: [2,3]
- The number of nodes in the list is in the range
-100 <= Node.val <= 100
- The list is guaranteed to be sorted in ascending order.
# 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): """ :type head: ListNode :rtype: ListNode """ dummy = ListNode(0); # construct a dummy node dummy.next = head pre = dummy # set up pre and cur pointers cur = head while cur: if cur.next and cur.val == cur.next.val: # loop until cur point to the last duplicates while cur and cur.next and cur.val == cur.next.val: cur = cur.next pre.next = cur.next # propose the next for pre # this will be verified by next line else: pre = pre.next cur = cur.next return dummy.next