LeetCode - 21 合并两个有序链表

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_head

        while list1 != None and list2 != None:
            if list1.val < list2.val:
                tail.next = list1
                list1 = list1.next
            else:
                tail.next = list2
                list2 = list2.next
            
            tail = tail.next
        
        tail.next = list1 if list1 else list2

        return 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;
};