leetcode刷题系列----模式2（Datastructure 链表）- 21：Merge Two Sorted Lists 合并两个有序链表-优快云博客

本文介绍了一种使用迭代法合并两个有序链表的方法，并提供了Python、C++、C#及Java四种语言的实现代码。该算法通过创建一个新链表并比较两个输入链表的节点值来决定插入顺序。

leetcode刷题系列----模式2（Datastructure 链表）- 21：Merge Two Sorted Lists 合并两个有序链表

Tips

更多题解请见本系列目录
使用迭代法解决。
链表类问题注意指针要随着迭代更新到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]:
        head = ListNode()
        point = head
        while list1 and list2:
            if list1.val < list2.val:
                point.next = list1
                list1 = list1.next
            else:
                point.next = list2 
                list2 = list2.next
            point = point.next
        point.next = list1 if list1 else list2 
        return head.next

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) {
        ListNode* head = new ListNode();
        ListNode* point = head;
        while(list1!=nullptr && list2!=nullptr)
        {
            if (list1->val < list2->val) {
                point->next = list1;
                list1 = list1->next;
            } else {
                point->next = list2;
                list2 = list2->next;
            }
            point = point->next;
        }
        
        point->next = list1==nullptr ? list2 : list1;
        return head->next;
    }
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode list1, ListNode list2) {
        ListNode head = new ListNode();
        ListNode point = head;
        while(list1!=null && list2!=null)
        {
            if(list1.val>list2.val)
            {
                point.next = list2;
                list2 = list2.next;
            }
            else
            {
                point.next = list1;
                list1 = list1.next;
            }
            point = point.next;
        }
        point.next = list1==null ? list2 : list1;
        return head.next;
    }
}

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode head = new ListNode();
        ListNode point = head;
        while(list1!=null && list2!=null)
        {
            if(list1.val>list2.val)
            {
                point.next = list2;
                list2 = list2.next;
            }
            else
            {
                point.next = list1;
                list1 = list1.next;
            }
            point = point.next;
        }
        point.next = list1==null ? list2 : list1;
        return head.next;
    }
}