LeetCode全集请参考：LeetCode Github 大全

题目

21. Merge Two Sorted Lists
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.

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:

The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both l1 and l2 are sorted in non-decreasing order.

递归解法，机器思考方式

倒序拼接，把最大的结果链表先返回，然后逐个拼接次大的。

/**
 * 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) {
        // check edge
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

遍历思维

1. 创建一个结果链表result；
2. 依次对比链表l1, l2,把小的node加入结果中，并指向下一个；
3. 记得首尾：把还不为空的链表加入到结果链表的尾部；
4. 开始的时候，可以检查边界，直接退出、

/**
 * 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) {
        // check edge
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        
        ListNode result = new ListNode(-1);
        ListNode dummy = result;
        
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                dummy.next = l1;
                l1 = l1.next;
            } else {
                dummy.next = l2;
                l2 = l2.next;
            }
            
            dummy = dummy.next;
        }
        
        // check not null;
        dummy.next = (l1 != null ? l1 : l2);
        
        
        return result.next;
    }
}