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题目来源

21. 合并两个有序链表 - 力扣（LeetCode）

题目描述

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

示例

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

题目解析

本题的要求是：需要使用 list1 和 list2 的节点来构成合并后的链表。

这里我们可以为合并后的链表创建一个虚拟头节点 dummy_head，之后按照下面图示逻辑进行：

C源码实现

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) {struct ListNode* dummy_head = (struct ListNode*)malloc(sizeof(struct ListNode));dummy_head->val = 0;dummy_head->next = NULL;struct ListNode* tail = dummy_head;while (list1 != NULL && list2 != NULL) {if (list1->val < list2->val) {tail->next = list1;list1 = list1->next;} else {tail->next = list2;list2 = list2->next;}tail = tail->next;}tail->next = list1 != NULL ? list1 : list2;return dummy_head->next;
}

C++源码实现

/*** 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* dummy_head = new ListNode(0, nullptr);ListNode* tail = dummy_head;while (list1 != nullptr && list2 != nullptr) {if (list1->val < list2->val) {tail->next = list1;list1 = list1->next;} else {tail->next = list2;list2 = list2->next;}tail = tail->next;}tail->next = list1 != nullptr ? list1 : list2;return dummy_head->next;}
};

Java源码实现

/*** 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 dummy_head = new ListNode(0, null);ListNode tail = dummy_head;while (list1 != null && list2 != null) {if (list1.val < list2.val) {tail.next = list1;list1 = list1.next;} else {tail.next = list2;list2 = list2.next;}tail = tail.next;}tail.next = list1 != null ? list1 : list2;return dummy_head.next;}
}

Python源码实现

# 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, list1, list2):""":type list1: Optional[ListNode]:type list2: Optional[ListNode]:rtype: Optional[ListNode]"""dummy_head = ListNode(0, None)tail = dummy_headwhile list1 != None and list2 != None:if list1.val < list2.val:tail.next = list1list1 = list1.nextelse:tail.next = list2list2 = list2.nexttail = tail.nexttail.next = list1 if list1 else list2return dummy_head.next

JavaScript源码实现

/*** Definition for singly-linked list.* function ListNode(val, next) {*     this.val = (val===undefined ? 0 : val)*     this.next = (next===undefined ? null : next)* }*/
/*** @param {ListNode} list1* @param {ListNode} list2* @return {ListNode}*/
var mergeTwoLists = function (list1, list2) {const dummy_head = new ListNode(0, null);let tail = dummy_head;while (list1 != null && list2 != null) {if (list1.val < list2.val) {tail.next = list1;list1 = list1.next;} else {tail.next = list2;list2 = list2.next;}tail = tail.next;}tail.next = list1 != null ? list1 : list2;return dummy_head.next;
};