分隔链表

1，程序简介

给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有小于 x 的节点都出现在大于或等于 x 的节点之前。
你应当保留两个分区中每个节点的初始相对位置。

示例 1：

输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]

示例 2：

输入：head = [2,1], x = 2
输出：[1,2]

提示：

链表中节点的数目在范围 [0, 200] 内
-100 <= Node.val <= 100
-200 <= x <= 200

2，程序代码

# -*- coding: utf-8 -*-
"""
Created on Mon Jan 10 21:42:17 2022
Function: 分隔链表
@author: 小梁aixj
"""
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None
class LinkList:
    def __init__(self):
        self.head=None
    def initList(self, data):
        self.head = ListNode(data[0])
        r=self.head
        p = self.head
        for i in data[1:]:
            node = ListNode(i)
            p.next = node
            p = p.next
        return r
    def    convert_list(self,head):
        ret = []
        if head == None:
            return
        node = head
        while node != None:
            ret.append(node.val)
            node = node.next
        return ret
class Solution(object):
    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        if head is None:
            return None
        less = lesshead = None
        last = pos = head
        while pos is not None:
            if pos.val < x:
                if lesshead is None:
                    lesshead = pos
                else:
                    less.next = pos
                less = pos
                if head == pos:
                    last = head = pos.next
                else:
                    last.next = pos.next
            else:
                last = pos
            pos = pos.next
        if lesshead is not None:
            less.next = head
        else:
            lesshead = head
        return lesshead
# %%
l = LinkList()
list1 = [1,4,3,2,5,2]
l1 = l.initList(list1)
x = 3
s = Solution()
print(l.convert_list(s.partition(l1, x)))#[1,2,2,4,3,5]