83. Remove Duplicates from Sorted List Leetcode Python
2015-02-04 23:56
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Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
定义一个pre 和cur
1.当二者不等的时候pre.next=cur pre=pre.next
2.当相等的时候pre.next=cur.next
3.每步都是cur=cur.next
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.next
cur=dummy.next
while cur:
if pre.val!=cur.val:
pre.next=cur
pre=pre.next
else:
pre.next=cur.next
cur=cur.next
return dummy.next
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
定义一个pre 和cur
1.当二者不等的时候pre.next=cur pre=pre.next
2.当相等的时候pre.next=cur.next
3.每步都是cur=cur.next
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.next
cur=dummy.next
while cur:
if pre.val!=cur.val:
pre.next=cur
pre=pre.next
else:
pre.next=cur.next
cur=cur.next
return dummy.next
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