（LeetCode）链表

0. 总结

链表问题可以用递归或者循环解决，相对来讲，循环解决更加直观，可以优先使用。

1. 反转链表

206. reverse-linked-list

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

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode

        笨办法
        1. 遍历链表，生成一个数组；
        2. 根据数组，倒序创建链表；

        聪明办法
        改指针方向
        """
        cur = head
        pre = None
        while cur is not None:
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next
        return pre

2. 合并两个有序链表

21. merge-two-sorted-lists

# 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 mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode

        笨办法：
        遍历两个链表，得到数组；将数组排好序；生成一个新的链表；

        聪明解法：
        递归
        """
        if l1 is None:
            return l2
        if l2 is None:
            return l1
        
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        if l1.val >= l2.val:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2