(Leet Code c++)Merge Two Sorted Lists

21. 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.

정렬된 ListNode를 합쳐서 리턴해야 한다.

l1과 l2가 NULL을 가리킬 때까지 반복하면서, answer 노드에 값을 추가하면 된다.

/**
 * 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 * answer = NULL;
        ListNode ** current = &answer;
        
        while(l1 != NULL || l2 != NULL){
            if(l1 == NULL){
                *(current) = new ListNode(l2->val);
                l2 = l2->next;
            }
            else if(l2 == NULL){
                *(current) = new ListNode(l1->val);
                l1 = l1->next;
            }
            else if(l1->val < l2->val){
                *(current) = new ListNode(l1->val);
                l1 = l1->next;
            }
            else{
                *(current) = new ListNode(l2->val);
                l2 = l2->next;
            }
            current = &((*current)->next);
        }
        
        return answer;
    }
};