python 合并两个有序链表

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/aBIT_Tu/article/details/83615261

参考了这个博主的答案（https://blog.csdn.net/gleam_/article/details/80149010），之前递归一直学的不是太好，但真的想通了以后写起来很神清气爽。

1. 递归的思路

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1==None and l2==None:
            return None
        if l1==None:
            return l2
        if l2==None:
            return l1
        if l1.val<=l2.val:
            l1.next=self.mergeTwoLists(l1.next,l2)
            return l1
        else:
            l2.next=self.mergeTwoLists(l1,l2.next)
            return l2

2. 非递归（这个是参看这个博主小哥的，https://blog.csdn.net/honeyaya/article/details/52915759）

# -*- 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
        if pHead1==None and pHead2==None:
            return None
        elif pHead1!=None and pHead2==None:
            return pHead1
        elif pHead1==None and pHead2!=None:
            return pHead2
        else:
            p1=pHead1
            p2=pHead2
            if p1.val<p2.val:
                newHead=p1
                p1=p1.next
            else:
                newHead=p2
                p2=p2.next
            p=newHead
            while p1!=None and p2!=None:
                if p1.val<p2.val:
                    p.next=p1
                    p=p.next
                    p1=p1.next
                else:
                    p.next=p2
                    p=p.next
                    p2=p2.next
            if p1!=None:
                while p1!=None:
                    p.next=p1
                    p1=p1.next
                    p=p.next
            if p2!=None:
                while p2!=None:
                    p.next=p2
                    p2=p2.next
                    p=p.next
            return newHead
！