# 21. Merge Two Sorted Lists https://leetcode.com/problems/merge-two-sorted-lists/ ### 我的想法 ```python class Solution: def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: curr1 = l1 curr2 = l2 temp = ListNode() while (curr1 and curr2) and(curr1.next and curr2.next): if curr1.val >= curr2.val: temp.next = curr1 curr1 = curr1.next else: temp.next = curr2 curr2 = curr2.next return temp.next ``` 思考的想法大概是 用一個新的ListNode，去串接答案 由curr1指向l1上的Node,curr2指向l2上的Node, 比較誰的值比較小；新的ListNode就接上(curr1,curr2)指到較小的那個Node ### 網友解法 ```python class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: head = sort_list = ListNode(0) while(l1 and l2): if (l1.val < l2.val): sort_list.next = l1 l1 = l1.next sort_list = sort_list.next elif (l1.val >= l2.val): sort_list.next = l2 l2 = l2.next sort_list = sort_list.next sort_list.next = l1 or l2 return head.next ``` 搞不懂為何sort_list在操作，最後是return head.next 在 ``` head = sort_list = ListNode(0) ``` 這行時，head , sort_list 被賦予同一個位子嗎? 所以sort_list後面做的事同時會接到head後面嗎? 目前只能這樣理解，但不曉得正確的觀念。 另外還有sort_list.next = l1 or l2這行 #### 亂想的過程 一開始的想法是有點混亂 第一個 if - elif 是在排除當其中是空的LinkedList時的狀況 第二個 if - elif 是看要從哪一串開始當Head 然後curr1：指向head後面的的node，curr2：指向另一條node * 當curr1指到的這個node的下一個值，大於curr2指到的值時 * 代表現在curr1.next要指到curr2目前這個node * 原本curr1後面的LinkedList 要讓curr2指標指過去 但這樣好想有點想得太複雜了 ```python # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: if l1 == None : return l2 elif l2 == None: return l1 if l1.val <=l2.val: head = l1 curr1 = head curr2 = l2 else : head = l2 curr2 = l1 curr1 = head while (curr1 and curr2) and(curr1.next and curr2.next): if curr1.next.val > curr2.val: #先例用一個暫時的Node把curr1.next後面那串接起來 temp = ListNode() temp.next = curr1.next curr1.next = curr2 #curr1.next接到curr2 curr1 = curr1.next #curr1往下移一格 curr2 = temp.next #curr2指到原本curr1.next那串 else: curr1=curr1.next return head ```