LeetCode:86. 分隔链表

1、题目描述

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

2、题解

2.1、解法一

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

class Solution:

    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        node = head
        prev = None
        stack = []

        ## 提取
        while node:
            if node.val < x:
                if not prev:
                    head = node.next
                else:
                    prev.next = node.next
                stack.append(node)
            else:
                prev = node
            node = node.next

        ## 插入
        cur_node = None
        while len(stack) >0:
            tmp = stack.pop(0)
            tmp.next = None
            if not cur_node:
                tmp.next = head
                head = tmp
            else:
                tmp.next = cur_node.next
                cur_node.next = tmp
            cur_node = tmp

        ## 获取
        n = head
        ret = []
        while n:
            # print(n.val)
            ret.append(n.val)
            n = n.next
            
        return ret

  

posted @ 2018-12-04 17:26  RobotsRising  阅读(160)  评论(0编辑  收藏  举报