题解 | #牛群分隔#

• 题目考察的知识点 : 链表的遍历和链表拼接操作
• 题目解答方法的文字分析：

1. 初始化小于和大于等于 x 的两个链表
2. 遍历原链表,按照 x 把节点分别接到两个链表尾部
3. 连接小链表在前,大链表在后,返回小链表头部

• 本题解析所用的编程语言：Python
• 完整且正确的编程代码

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param x int整型 
# @return ListNode类
#
class Solution:
    def cow_partition(self , head: ListNode, x: int) -> ListNode:
        small = ListNode(0)
        large = ListNode(0)

        smallHead, largeHead = small, large 
        while head:
            if head.val < x:
                small.next = head
                small = small.next
            else:
                large.next = head
                large = large.next
    
            head = head.next
  
        large.next = None
        small.next = largeHead.next
  
        return smallHead.next