Given a sorted linked list, delete all nodes that have duplicate
numbers, leaving only distinct numbers from the original list.
https://leetcode.com/problems...
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def deleteDuplicates(self, head: ListNode) -> ListNode: # Two nodes: one to keep head, one to keep tail. ahead = phead = ListNode("null") # Two pointers: one to move ahead, one to monitor duplicates. p = None c = head # When move to end, terminate. while c: # At the beginning, simple move one step forward. if p == None: p = c # If p and c have different values, need to determine why. elif p.val != c.val: # If p and c are neighbors, p must be unique value. if p.next == c: phead.next = p phead = phead.next phead.next = None # p gets one step forward. p = c # If p and c are not neighbors, ignore nodes with p.val. else: p = c # c moves one step anyway. c = c.next # If p is not None (input not empty) and has next, collect last element. if p != None and p.next == None: phead.next = p return ahead.next
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