链表-（合并两个有序链表）的算法

# -*- coding:utf-8 -*-

""" 
Author: leadingme
Mail:[email protected]
MyWebsite:leadingme.top
"""

# 合并两个有序链表
"""
    算法要求:
        将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表所有节点组成的。
    示例:
        输入: 1->2->4, 1->3->4
        输出: 1->1->2->3->4->4
"""

def mergeTwoLists(l1, l2):
    """

    :type l1: LinkNode
    :param l2: LinkNode
    :return:
    """
    if l1 is None and l2 is None: # 两条链表为空链表
        return None
    fakeNode = LinkNode(0) # 创建一个val为0的新节点，用来作为新链表的头节点
    pointer = fakeNode
    while l1 is not None or l2 is not None:
        if l1 is None: # l1链表为空, l2链表为非空, 遍历完整个l2
            pointer.next = l2
            l2 = l2.next
            pointer = pointer.next
        elif l2 is None:
            pointer.next = l1
            l1 = l1.next
            pointer = pointer.next
        else:
            if l1.val <= l2.val:  # 如果两个链表都不为空，则比较l1.val和l2.val,较小的放前面
                pointer.next = l1
                l1 = l1.next
                pointer = pointer.next
            else:
                pointer.next = l2
                l2 = l2.next
                pointer = pointer.next
        return fakeNode.next