21. Merge two ordered linked lists

21. Merge two ordered lists

The difficulty is simple 1983

Merge two ascending linked lists into a new   Ascending   Link list and return . The new linked list is made up of all the nodes of the given two linked 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]

Tips ：

• The range of the number of nodes in the two linked lists is  [0, 50]
• -100 <= Node.val <= 100
• l1  and  l2  All according to   Non decreasing order   array

/**
 * 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) {

    }
};

#include <iostream>

using namespace std;

/// Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

/// Iterative
/// Time Complexity: O(len(l1) + len(l2))
/// Space Complexity: O(1)
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

        ListNode* dummyHead = new ListNode(-1);
        ListNode* p = dummyHead;
        ListNode* l1p = l1;
        ListNode* l2p = l2;
        while(l1p != NULL && l2p != NULL){
            if(l1p->val < l2p->val){
                p->next = l1p;
                l1p = l1p->next;
            }
            else{
                p->next = l2p;
                l2p = l2p->next;
            }

            p = p->next;
        }

        if(l1p != NULL)
            p->next = l1p;
        else
            p->next = l2p;

        ListNode* ret = dummyHead->next;
        dummyHead->next = NULL;
        delete dummyHead;

        return ret;
    }
};

int main() {

    return 0;
}