Leetcode 21. Merge Two Sorted Lists

Description:Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists.

Link: https://leetcode.com/problems/merge-two-sorted-lists/

What you need to know: basic functions of linked list, such as insert, delete and add (refer to https://blog.csdn.net/qq_34364995/article/details/80787090).

想法：当我们对链表的操作还不是很熟悉时，不妨先用普通的list来构思解题思路，然后把全部的function换成链表的操作。所以如果是对 two sorted list排序：

l3 = []
while j < len(l2):
    while i < len(l1) and l1[i] <= l2[j]:
        l3.append(l1[i])
        i += 1
    l3.append(l2[j])
    j += 1
return l3

全部换成 linked list:

head = ListNode()
p = head
while l2:
    while l1 and (l1.val <= l2.val):
        p.next = l1
        p = p.next
        l1 = l1.next
    p.next = l2
    p = p.next
    l2 = l2.next
return head.next

考虑特殊情况: (1) one of l1 and l2 is None, and (2) l1 = [2], l2 = [1], 需要将剩余的尾部连上去，final version:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        head = ListNode()
        p = head
        if l1 is None: return l2
        if l2 is None: return l1
        
        while l2:
            while l1 and (l1.val <= l2.val):
                p.next = l1
                p = p.next
                l1 = l1.next
            p.next = l2
            p = p.next
            l2 = l2.next
        if l1:
            p.next = l1
        if l2:
            p.next = l2
        return head.next

日期：2020-11-13 (Time and efforts are always needed. Closer to the deadline.)