[Lintcode]96. Partition List/[Leetcode]86. Partition List

96. Partition List/86. Partition List

• 本题难度: Easy/Medium
• Topic: Linked List

Description

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
Example 1:
Input: list = null, x = 0
Output: null

Explanation:
The empty list Satisfy the conditions by itself.

Example 2:
Input: list = 1->4->3->2->5->2->null, x = 3
Output: 1->2->2->4->3->5->null

Explanation:  
keep the original relative order of the nodes in each of the two partitions.

我的代码

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""

class Solution:
    """
    @param head: The first node of linked list
    @param x: An integer
    @return: A ListNode
    """
    def partition(self, head, x):
        # write your code here
        if head is None:
            return head
        l1 = ListNode(0)
        l2 = ListNode(0)
        p1, p2 = l1, l2
        while(head):
            if head.val<x:
                l1.next = head
                l1 = l1.next
            else:
                l2.next = head
                l2 = l2.next
            head = head.next
        l1.next = None#防止结尾为错误
        l2.next = None
        if p1.next is None: #注意是p1而不是l1
            return p2.next
        else:
            l1.next = p2.next# 需要考虑p1.next is None的情况
            return p1.next

思路

分成两个数组，最后合到一起

1. null
   2.head.next == null

• 时间复杂度 O(n)
• 出错

1. 当出现xx.next =yyy的时候要考虑到xx.next是否存在
2. 最后一个值可能小于x，但是因为是直接接head，所以可能会残余。所以最后需要加上结尾为null
3. 当前位置和开头位置的表示搞混了