力扣刷题-热题100题-第27题（c++、python）

21. 合并两个有序链表 - 力扣（LeetCode）https://leetcode.cn/problems/merge-two-sorted-lists/description/?envType=study-plan-v2&envId=top-100-liked

常规法

创建一个新链表，遍历list1与list2，将新链表指向list1与list2中更小的那个直至结束。

//c++
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) 
    {
        if(list1==nullptr||list2==nullptr)  return list1?list1:list2;
        ListNode* ans=new ListNode(-1);
        ListNode* aa=ans;
        while(list1 && list2)
        {
            if(list1->val<=list2->val)
            {
                aa->next=list1;
                list1=list1->next;
            }
            else
            {
                aa->next=list2;
                list2=list2->next;
            }
            aa=aa->next;
        }
        aa->next=list1?list1:list2;
        return ans->next;
    }
};

#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, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if list1==None or list2==None:
            return list1 if list1 else list2
        ans=ListNode()
        aa=ans
        while list1 and list2:
            if list1.val<=list2.val:
                aa.next=list1
                list1=list1.next
            else:
                aa.next=list2
                list2=list2.next
            aa=aa.next
        aa.next=list1 if list1 else list2
        return ans.next

递归法

进入递归，哪个值更小，就使更小值指向进入对应链表的下一个指针的递归并返回。

//c++
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) 
    {
        if(list1==nullptr)  return list2;
        else if(list2==nullptr)    return list1;
        else if(list1->val<=list2->val)
        {
            list1->next=mergeTwoLists(list1->next,list2);
            return list1;
        }
        else
        {
            list2->next=mergeTwoLists(list1,list2->next);
            return list2;
        }
    }
};

#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, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if list1 is None:
            return list2
        elif list2 is None:
            return list1
        elif list1.val<=list2.val:
            list1.next=self.mergeTwoLists(list1.next,list2)
            return list1
        else:
            list2.next=self.mergeTwoLists(list1,list2.next)
            return list2