82. Remove Duplicates from Sorted List II Leetcode Python

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

82. Remove Duplicates from Sorted List II Leetcode Python

原文：http://blog.csdn.net/hyperbolechi/article/details/43526043