21. 合并两个有序链表

21. 合并两个有序链表

难度简单1983

将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

/**
 * 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;
}