21. Merge Two Sorted Lists

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21. Merge Two Sorted Lists

Description

Difficulty: Easy

Related Topics: Linked List, Recursion

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:

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

Example 2:

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

Example 3:

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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.

Hints/Notes

  • 2023/07/31
  • Dummy pointer
  • After one list is exhausted, we can just link the remaining of other list to the new list. The corner case that one list is empty is covered with this handling.
  • 0x3F’s solution(checked)

Solution

Language: C++

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