[LeetCode January Challange] Day 4 - Merge Two Sorted Lists

Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists.

Example 1:

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: l1 = [], l2 = []
Output: []

Example 3:

Input: l1 = [], l2 = [0]
Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both l1 and l2 are sorted in non-decreasing order.

---

Solution

Time complexity : O(n)
Space complexity : O(n)

/**
 * 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* l1, ListNode* l2) {
        ListNode* dummy = new ListNode();
        ListNode* cur = dummy;
        
        while (l1 && l2) {
            if (l1->val < l2->val)
                append_node(&cur, &l1);
            else
                append_node(&cur, &l2);
        }
        
        while (l1) append_node(&cur, &l1);
        while (l2) append_node(&cur, &l2);
        
        return dummy->next;
    }
    
    void append_node(ListNode** cur, ListNode** node) {
        (*cur)->next = new ListNode((*node)->val);
        *cur = (*cur)->next;
        *node = (*node)->next;
    }
};

當 l1、l2 都還有東西時，互相比較，取小者。

當l1或是l2其中一方跑完，則取另一方所剩。