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题目：
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
代码：oj在线测试通过 Runtime: 208 ms
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 # @param two ListNodes 9 # @return a ListNode 10 def mergeTwoLists(self, l1, l2): 11 if l1 is None: 12 return l2 13 if l2 is None: 14 return l1 15 16 dummyhead = ListNode(0) 17 dummyhead.next = None 18 p = dummyhead 19 while l1 is not None and l2 is not None: 20 if l1.val > l2.val: 21 p.next = l2 22 l2 = l2.next 23 else: 24 p.next = l1 25 l1 = l1.next 26 p = p.next 27 if l1 is not None: 28 p.next = l1 29 else: 30 p.next = l2 31 return dummyhead.next
思路：
虚表头dummyhead设定好了就不要动，设定一个p=hummyhead，然后从处理p.next开始；最后返回hummyhead就可以获得正确的结果
两个表的指针一步步移动，并通过比较val的大小决定哪个插入p.next
注意，每次执行完判断，要移动指针p=p.next
leetcode 【 Merge Two Sorted Lists 】 python 实现
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4186905.html
