leetcode86

class Solution:
    def constructLink(self,lists):
        n = len(lists)
        if n == 0:
            return None
        if n == 1:
            return ListNode(lists[0])
        
        head = ListNode(lists[-1])
        for i in range(n-2,-1,-1):
            cur = ListNode(lists[i])
            cur.next = head 
            head = cur
        return head
    
    def partition(self, head: ListNode, x: int) -> ListNode:
        lists = []
        while head != None:
            lists.append(head.val)
            head = head.next
        before = []
        after = []
        for i in range(len(lists)):
            if lists[i] < x:
                before.append(lists[i])
            else:
                after.append(lists[i])
        return self.constructLink(before + after)

将链表中的节点，按照x值分别存储到两个集合中，再用重新拼接的集合建立链表。