LeetCode

Merge Two Sorted Lists

2013.12.22 03:24

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

Solution:

　　This is a FAQ in IT interview. Make sure you don't new any node, and watch out for any special cases like NULL pointer or what.

　　Time complexity is O(m + n), where m and n are the lengths of two lists. Space complexity is O(1).

Accepted code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(l1 == nullptr){
            return l2;
        }else if(l2 == nullptr){
            return l1;
        }
        
        ListNode *l3, *head;
        
        if(l1->val <= l2->val){
            l3 = l1;
            l1 = l1->next;
        }else{
            l3 = l2;
            l2 = l2->next;
        }
        l3->next = nullptr;
        head = l3;
        
        while(l1 != nullptr && l2 != nullptr){
            if(l1->val <= l2->val){
                l3->next = l1;
                l1 = l1->next;
            }else{
                l3->next = l2;
                l2 = l2->next;
            }
            l3 = l3->next;
            l3->next = nullptr;
        }
        
        while(l1 != nullptr){
            l3->next = l1;
            l1 = l1->next;
            l3 = l3->next;
            l3->next = nullptr;
        }
        while(l2 != nullptr){
            l3->next = l2;
            l2 = l2->next;
            l3 = l3->next;
            l3->next = nullptr;
        }
        
        return head;
    }
};