Description
输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
Solutions
可以用迭代的方法搞定：
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回合并后列表
def Merge(self, pHead1, pHead2):
pRes = None
pNode = None
while pHead1 and pHead2:
pNext = None
if pHead1.val <= pHead2.val:
pNext = pHead1
pHead1 = pHead1.next
else:
pNext = pHead2
pHead2 = pHead2.next
if pNode is None or pRes is None:
pRes = pNext
pNode = pNext
continue
pNode.next = pNext
pNode = pNode.next
if pHead1 or pHead2:
if pNode is not None:
pNode.next = pHead1 or pHead2
else:
pRes = pHead1 or pHead2
return pRes
当然，利用迭代方式很明显不是很优雅，可以尝试使用递归的方式实现：
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回合并后列表
def Merge(self, pHead1, pHead2):
if pHead1 is None :
return pHead2
elif pHead2 is None:
return pHead1
pNode = None
if pHead1.val <= pHead2.val:
pNode = pHead1
pNode.next = self.Merge(pHead1.next, pHead2)
else:
pNode = pHead2
pNode.next = self.Merge(pHead1, pHead2.next)
return pNode
要是要首先搞清楚递归的过程再去写代码，别写代码以猜的心态，交给上帝去调试？