合并两个有序链表

题目描述：

合并两个已排序的链表，并将其作为一个新列表返回。新列表应该通过拼接前两个列表的节点来完成。 

解题思路：

构造一个新的链表空间，利用链表的next依次组合，不难，就是生疏了忘了链表看了下。

代码：

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        #newnode = ListNode(None)  # define the res ListNode
        if l1 ==None and l2 == None:  # l1 and l2 both are none
            return None
        if l1 == None:  # l1 is none
            return l2
        if l2 == None:  # l2 is none
            return l1
        #  initialize the res ListNode
        if l1.val >= l2.val:
            newnode = l2
            l2 = l2.next
        else:
            newnode = l1
            l1 = l1.next
        res = newnode
        while l1 != None and l2 != None:  # l1 and l2 record in the res
            if l1.val <= l2.val:
                res.next = l1
                res = res.next
                l1 = l1.next
            else:
                res.next = l2
                res = res.next
                l2 = l2.next
        if l1 != None:  # l1 is not none but l2 is none
            res.next = l1
            res = res.next
            l1 = l1.next
        if l2 != None:  # l2 is not none but l1 is none
            res.next = l2
            res = res.next
            l2 = l2.next
        return newnode

'''进行了简化'''

代码：

# -*- 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):
        # write code here
        res = head = ListNode(0)    
        while pHead1 and pHead2:
            if pHead1.val > pHead2.val:
                head.next = pHead2
                pHead2 = pHead2.next
            else:
                head.next = pHead1
                pHead1 = pHead1.next
            head = head.next   
        if pHead1:
            head.next = pHead1
        if pHead2:
            head.next = pHead2
        return res.next

这道题在剑指offer上也出现了，看到了新的解法，就是按照递归的方法进行解决问题。

代码：

# -*- 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 not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        pHead = None
        if pHead1.val <= pHead2.val:
            pHead = pHead1
            pHead.next = self.Merge(pHead1.next, pHead2)
        else:
            pHead = pHead2
            pHead.next = self.Merge(pHead1, pHead2.next)
        return pHead