【leetcode】86. 分隔链表

文章目录

- 题目
- 题解
- - [1. 先找大于x的节点，然后进行插入](#1. 先找大于x的节点，然后进行插入)
  - [2. 小链表，大链表](#2. 小链表，大链表)

题目

86. 分隔链表

给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有小于 x 的节点都出现在大于或等于 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。

题解

1. 先找大于x的节点，然后进行插入

python 复制代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def partition(self, head, x):
        """
        :type head: Optional[ListNode]
        :type x: int
        :rtype: Optional[ListNode]
        """
        dummy = ListNode(next=head)
        cur = dummy
        left = dummy
        right = None
        while cur and cur.next:
            if cur.next.val >= x:
                left = cur
                right = cur.next
                break
            cur = cur.next
        cur = right

        while cur and cur.next:
            if cur.next.val < x:
                tmp = cur.next
                cur.next = cur.next.next
                left.next = tmp
                tmp.next = right
                left = tmp
            else:
                cur = cur.next
        return dummy.next

2. 小链表，大链表

python 复制代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def partition(self, head, x):
        """
        :type head: Optional[ListNode]
        :type x: int
        :rtype: Optional[ListNode]
        """
        small_dummy = ListNode(0)
        big_dummy = ListNode(0)

        small = small_dummy
        big = big_dummy

        while head:
            if head.val < x:
                small.next = head
                small = small.next
            else:
                big.next = head
                big = big.next
            head = head.next
        small.next = big_dummy.next
        big.next = None
        return small_dummy.next