题目描述

解题思路

• 遍历l1，l2比较当前节点值得的大小；
• 将较小值的节点添加至新链表中；
• 将对应链表的节点向后移一位。

代码实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        ListNode l3 = new ListNode(0);
        ListNode res = l3;

        while(l1 != null || l2 != null){
            if(l1 == null){
                l3.next = l2;
                break;
            }
            if(l2 == null){
                l3.next = l1;
                break;
            }

            if(l1.val > l2.val){
                l3.next = l2;
                l2 = l2.next;
            }else{
                l3.next = l1;
                l1 = l1.next;
            }

            l3 = l3.next;
        }

        return res.next;
    }
}