题解 | #链表# #删除有序链表中重复的元素-I#

#coding:utf-8
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @return ListNode类
#
class ListNode:
     def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def print_node_list(self, node):
        ret = []
        if node == None:
            pass 
        while node != None:
            ret.append(node.val)
            node = node.next 
        print ("Ret is: ", ret)
        return  
    
    def deleteDuplicates(self , head ):
        # write code here
        ##init 
        if head == None:
            return None
        if head.next == None:
            return head
        dummy = ListNode(-1)
        dummy.next = head
        cur = head
        next_nd = cur.next
        ##process
        while cur != None:
            next_nd = cur.next #注意：1、上面循环不变量应该用cur 2、next应该在这里取
            cur_val = cur.val 
            #next_val = next_nd.val
            while next_nd != None and next_nd.val == cur_val:
                next_nd = next_nd.next
            cur.next = next_nd
            cur = next_nd
            
        #self.print_node_list(dummy.next) ##Debug使用

        return dummy.next