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Merge Two Sorted Lists

Tags : linkedlist, leetcode, cpp, easy

Problem Statement - link #

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.

Examples #

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 #

Solutions #

/**
 * 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) {}
 * };
 */
/**
 * 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) {
         if(l1 == NULL) return l2;
         if(l2 == NULL) return l1;
        
         ListNode* left = NULL, *right = NULL;
        
         if(l1->val <= l2->val){
             left = right = l1;
             l1 = l1->next;
         } 
         else{
             left = right = l2;
             l2 = l2->next;
         }
        
        while( l1!=NULL && l2!=NULL){
            if(l1->val <= l2->val){
                right->next = l1;
                right = right->next;
                l1 = l1->next;
            }else{
                right->next = l2;
                right = right->next;
                l2 = l2->next;
            }
        }
        
        if(l2 == NULL)
            right->next = l1;
        else
            right->next = l2;
        
        return left;
    }
};