python----兩個有序連結清單的合并

第一種方法：非遞歸

#defination of ListNode
class ListNode:
    def __init__(self,x):
        self.val=x
        self.next=None
class Solution:
    def ListNodeMerge(self,head1,head2):
        temp=ListNode(0)
        preList=temp
        if head1 is None and head2 is None:
            return None
        while head1 is not None and head2 is not None:
            if head1.val<head2.val:
                preList.next=head1
                head1=head1.next
            else:
                preList.next=head2
                head2=head2.next
            preList=preList.next
        if head1 is not None:
            preList.next=head1
        elif head2 is not None:
            preList.next=head2
        return temp.next

if __name__=="__main__":
    head1=ListNode(1)
    head1.next=ListNode(3)
    head2=ListNode(1)
    head2.next=ListNode(2)
    su=Solution()
    res=su.ListNodeMerge(head1,head2)
    result=""
    while res is not None:
        result=result+str(res.val)
        res=res.next
    print(result)
           

第二種方法：遞歸

#defination of ListNode
class ListNode:
    def __init__(self,x):
        self.val=x
        self.next=None
class Solution:
    def ListNodeMerge(self,head1,head2):
        if head1==None and head2==None:
            return None
        if head1==None:
            return head2
        if head2==None:
            return head1
        if head1.val<head2.val:
            head1.next=self.ListNodeMerge(head1.next,head2)
            return head1
        else:
            head2.next=self.ListNodeMerge(head1,head2.next)
            return head2

if __name__=="__main__":
    head1=ListNode(1)
    head1.next=ListNode(3)
    head2=ListNode(1)
    head2.next=ListNode(2)
    su=Solution()
    res=su.ListNodeMerge(head1,head2)
    result=""
    while res is not None:
        result=result+str(res.val)
        res=res.next
    print(result)