力扣——Partition List(分隔链表) python实现
题目描述:
中文:
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。
你应当保留两个分区中每个节点的初始相对位置。
示例:
输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5
英文:
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
# 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 """ head1 = ListNode(0) head2 = ListNode(0) Tmp = head phead1 = head1 phead2 = head2 while Tmp: if Tmp.val < x: phead1.next = Tmp Tmp = Tmp.next phead1 = phead1.next phead1.next = None else: phead2.next = Tmp Tmp = Tmp.next phead2 = phead2.next phead2.next = None phead1.next = head2.next head = head1.next return head
来源来源:力扣
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