leetcode 83& 86

思路：两个指针，一个是当前指针cur，一个是当前指针的下一个nex。如果cur.val
等于nex.val，那么cur.next = nex.next，nex向后移动为nex.next，当前指针cur不变，继续循环；如果cur.val不等于nex.val，那么cur和nex都向后移动一位，继续循环。

# 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':
        if head == None:
            return head
        cur = head
        nex = cur.next
        while nex:
            if cur.val == nex.val:
                cur.next = nex.next
                nex = nex.next
 
   
            else:
                cur = nex
                nex = nex.next
        return head 
                

思路：重新创建两个聊表，其中一个存储小于x的数，另一个存储大于x的数。创建链表的过程中，小于x的链表当前指针用less_point表示，大于x的链表当前指针用more_point表示。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def partition(self, head: 'ListNode', x: 'int') -> 'ListNode':
        smaller_dummy = ListNode(-1)
        less_point = smaller_dummy
        
        bigger_dummy = ListNode(-1)
        more_point = bigger_dummy
        
        cur = head
        
        while cur:
            if cur.val < x: 
                less_point.next = cur
                less_point = cur
                
            else:
                more_point.next = cur
                more_point = cur
            
            cur = cur.next
        
        less_point.next = bigger_dummy.next
        more_point.next = None
        return smaller_dummy.next