Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
Example:
Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。
但注意两个分区中每个节点的初始相对位置保持不变。
坑：空链表，x小于链表中所有值或大于链表中所有值。
思路：定义两个指针。ori指针始终在原链表中操作（<x的部分），new指针指向(>x的部分)。所以无需创建新结点。遍历完成后将两个链表合并即可。
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 def partition(self, head: ListNode, x: int) -> ListNode: 9 if not head: 10 return None 11 new = ListNode(0) # 定义一个新头结点，用于存放所有>x的数 12 new.next = None 13 recode_new = new 14 15 ori = ListNode(0) # 原始结点，保存所有原链表中<x的数 16 ori.next = head 17 recore_ori = ori 18 while ori and ori.next: 19 if ori.next.val<x: # 如果<x就接着往下走 20 ori = ori.next 21 else: 22 new.next = ori.next # 否则加入到new链表中 23 new = new.next 24 ori.next = ori.next.next 25 new.next = None 26 ori.next = recode_new.next # 合并两个链表 27 28 return recore_ori.next
验证：
1 # 新建链表1 2 listnode1 = ListNode_handle(None) 3 s1 = [666,4,3,2,58,5,78,2,634,42,36,244,35,41,33,32,23,266] 4 for i in s1: 5 listnode1.add(i) 6 listnode1.print_node(listnode1.head) 7 8 s = Solution() 9 head = s.partition(listnode1.head,137) 10 listnode1.print_node(head)
666 4 3 2 58 5 78 2 634 42 36 244 35 41 33 32 23 266
4 3 2 58 5 78 2 42 36 35 41 33 32 23 666 634 244 266
原文：https://www.cnblogs.com/king-lps/p/10664144.html