[leetcode]Partition List @ Python

原題地址：https://oj.leetcode.com/problems/partition-list/

題意：

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

解題思路：解決鏈表問題時，最好加一個頭結點，問題會比較好解決。對這道題來說，創建兩個頭結點head1和head2，head1這條鏈表是小於x值的節點的鏈表，head2鏈表是大於等於x值的節點的鏈表，然后將head2鏈表鏈接到head鏈表的尾部即可。

代碼：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @param x, an integer
    # @return a ListNode
    def partition(self, head, x):
        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