leetcode之Partition List

本题采用的是分离了2类链表，最后合在一起的方法，需要注意的是容易产生环。需要最后处理一下。代码如下：

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

class Solution(object):
    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        if not head:
            return head
        if not head.next:
            return head
        a = ListNode(1)
        b = ListNode(2)
        a1 = a
        b1 = b
        while head:
            if head.val < x:
                a1.next = head
                a1 = a1.next
                head = head.next
            else:
                b1.next = head
                b1 = b1.next
                head = head.next
            if not head:
                a1.next = None
                b1.next = None
        if not a.next:
            return b.next
        else:
            a1.next = b.next
            return a.next