LeetCode 86 Partition List (Python详解及实现)

【题目】

Given a linked list and a value x,partition it such that all nodes less than x come before nodes greater than orequal to x.

 

You should preserve the original relativeorder of the nodes in each of the two partitions.

 

For example,

Given 1->4->3->2->5->2 and x= 3,

return 1->2->2->4->3->5.

 

给一个链表，可一个定值x，在保证链表顺序不变的前提下，将小于x的放在x前面，大于等于x的放其后。

 

【思路】

 

最简单的是创建两个头结点head1和head2；head1这条链表是小于x值的节点的链表，head2链表是大于等于x值的节点的链表，最后将head2链表链接到head链表的尾部即可。

 

【Python实现】

# -*- coding: utf-8 -*-

"""

Created on Wed Aug  9 16:41:48 2017

 

@author: Administrator

"""

 

# 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)#新的头结点，用于存放小于x的数

       head2 = ListNode(0)#新的头结点，用于存放大于等于x的数

       p1 = head1

       p2 = head2

       curr = head

       while curr:

           if curr.val < x:

                p1.next = curr

                curr = curr.next

                p1 = p1.next

                p1.next = None

           else:

                p2.next = curr

                curr = curr.next

                p2 = p2.next

                p2.next = None

       p1.next = head2.next

       head = head1.next

       curr = head

       #打印新链表

       while curr:

           print(curr.val)

           curr = curr.next

       return head

   

if __name__ == '__main__':

    S= Solution()

   l1 = ListNode(1)

   l2 = ListNode(4)

    l3= ListNode(3)

   l4 = ListNode(2)

   l5 = ListNode(5)

   l6 = ListNode(2)

   head = l1

   l1.next = l2

   l2.next = l3

   l3.next = l4

   l4.next = l5

   l5.next = l6

   l6.next = None

   S.partition(head,3)