【leetcode】86. Partition List

题目如下：

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

解题思路：我的方法和【leetcode】328. Odd Even Linked List 一样，使用less 和 great 两个指针，分别指向小于和等于或大于的节点，最后再把less的尾节点指向great的头结点即可。

代码如下：

# Definition for singly-linked list.
class ListNode(object):def __init__(self, x):self.val = xself.next = Noneclass Solution(object):def partition(self, head, x):""":type head: ListNode:type x: int:rtype: ListNode"""lessHead = less = greatHead = great = Nonecurrent = headwhile current != None:tmp = current.nextcurrent.next = Noneif current.val < x:if lessHead == None:lessHead = currentless = lessHeadelse:less.next = currentless = less.nextelse:if greatHead == None:greatHead = great = currentelse:great.next = currentgreat = great.nextcurrent = tmpif less != None:less.next = greatHeadelse:lessHead = greatHeadreturn lessHead

转载于:https://www.cnblogs.com/seyjs/p/10194292.html

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