Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
这题的做法是用两个指针pre cur
1.dummy.next= head
2.pre=dummy cur=dummy,next
3.当 指针移到 pre.next 和cur.next 不等的位置 将这个位置加入到 pre.next中去
4.否者就跳过cur
一次遍历 时间是O(n)
代码如下
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param head, a ListNode # @return a ListNode def deleteDuplicates(self, head): if head==None or head.next==None: return head dummy=ListNode(0) dummy.next=head pre=dummy cur=dummy.next while cur!=None: while cur.next and cur.next.val==pre.next.val: cur=cur.next if pre.next==cur: pre=pre.next else: pre.next=cur.next cur=cur.next return dummy.next
时间: 2024-11-08 09:22:09