Lintcode 合并两个排序的链表

来源：互联网发布：远程桌面控制软件编辑：程序博客网时间：2024/05/18 03:06

将两个排序链表合并为一个新的排序链表
样例
给出 1->3->8->11->15->null，2->null，返回 1->2->3->8->11->15->null。
递归实现：

"""Definition of ListNodeclass ListNode(object):    def __init__(self, val, next=None):        self.val = val        self.next = next"""class Solution:    """    @param two ListNodes    @return a ListNode    """    def mergeTwoLists(self, l1, l2):        # write your code here        if l1 == None:            return l2        if l2 == None:            return l1        mergeHead = None        if l1.val <= l2.val:            mergeHead = l1            mergeHead.next = self.mergeTwoLists(l1.next,l2)        elif l2.val < l1.val:            mergeHead = l2            mergeHead.next = self.mergeTwoLists(l1,l2.next)        return mergeHead

非递归实现：

"""Definition of ListNodeclass ListNode(object):    def __init__(self, val, next=None):        self.val = val        self.next = next"""class Solution:    """    @param two ListNodes    @return a ListNode    """    def mergeTwoLists(self, l1, l2):        # write your code here        if l1 == None:            return l2        if l2 == None:            return l1        ptr1 = l1        ptr2 = l2        l = ListNode(0)        current = l        while(ptr1!= None and ptr2 != None):            if ptr1.val < ptr2.val:                current.next = ptr1                ptr1 = ptr1.next            elif ptr1.val > ptr2.val:                current.next = ptr2                ptr2 = ptr2.next            else:                current.next = ptr1                ptr1 = ptr1.next                current = current.next                current.next = ptr2                ptr2 = ptr2.next            current = current.next        if ptr1 == None:            current.next = ptr2        else:            current.next = ptr1        l = l.next        return l

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