LeetCode 21 - Merge Two Sorted Lists

Difficulty: easy

Problem Description

English (Merge Two Sorted Lists)

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

graph LR

node11((1))
node12((2))
node13((4))
node11 -----> node12 -----> node13

node21((1))
node22((3))
node23((4))
node21 -------> node22 -----> node23

node31((1))
node32((1))
node33((2))
node34((3))
node35((4))
node36((4))
node31 ---> node32 ---> node33 ---> node34 ---> node35 ---> node36

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

Chinese (合并两个有序链表)

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 1：

graph LR

node11((1))
node12((2))
node13((4))
node11 -----> node12 -----> node13

node21((1))
node22((3))
node23((4))
node21 -------> node22 -----> node23

node31((1))
node32((1))
node33((2))
node34((3))
node35((4))
node36((4))
node31 ---> node32 ---> node33 ---> node34 ---> node35 ---> node36

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

Solution

/**
 * 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 = new ListNode;

        ListNode* last = dummy;
        while (list1 != nullptr && list2 != nullptr) {
            if (list1->val <= list2->val) {
                last->next = list1;
                list1 = list1->next;
            } else {
                last->next = list2;
                list2 = list2->next;
            }

            last = last->next;
        }

        if (list1 != nullptr) {
            last->next = list1;
        }
        if (list2 != nullptr) {
            last->next = list2;
        }

        ListNode* head = dummy->next;
        delete dummy;
        return head;
    }
};