【LeetCode with Python】 Partition List

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原题页面： https://oj.leetcode.com/problems/partition-list/
题目类型：
难度评价：★
本文地址： http://blog.youkuaiyun.com/nerv3x3/article/details/39453367

Given a linked list and a value x, partition it such that all nodes less thanx 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.

class Solution:
    # @param head, a ListNode
    # @param x, an integer
    # @return a ListNode
    def partition(self, head, x):
        if None == head or None == head.next:
            return head
        new_head = ListNode(-1)
        new_head.next = head
        tail = head
        while None != tail.next:
            tail = tail.next
        cur = head
        parent = new_head
        l_pos = new_head

        while tail != parent and None != cur:
            if cur.val >= x:
                break
            parent = parent.next
            cur =  cur.next
            l_pos = l_pos.next
        parent =  parent.next
        if None != cur:
            cur = cur.next
        #l_pos = l_pos.next

        while tail != parent and None != cur:
            if cur.val < x:
                parent.next = cur.next
                cur.next = l_pos.next
                l_pos.next = cur
                l_pos = l_pos.next
                cur = parent.next
            else:
                parent = parent.next
                cur =  cur.next
        return new_head.next