86. 分隔链表

思路：

1、遍历原链表，分别找到小于x的节点与大于等于x的节点组成的两个链表small和big;

2、拼接small和big，small在前。

class Solution(object):
    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        if not head:
            return head
        # 小于x的节点，ans.next是返回值
        small = ans = ListNode(-1)
        # 不小于x的节点
        big = head_big = ListNode(-1)
        # 遍历指针
        cur = head
        while cur:
            if cur.val < x:
                small.next = cur
                small = small.next
            else:
                big.next = cur
                big = big.next
            cur = cur.next
        # 将两个链表段拼接，小于x的在前，不小于x的在后
        big.next = None
        small.next = head_big.next
        return ans.next