链表-面试题25. 合并两个排序的链表

解题思路：迭代

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur=dummy;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                cur.next=l1;
                l1=l1.next;
            }
            else{
                cur.next=l2;
                l2=l2.next;
            }
            cur=cur.next;
        }
        cur.next=l1==null?l2:l1;
        return dummy.next;
    }
}

 解题思路二：递归

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        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;
        }
    }
}

参考博客链接：

https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof/solution/he-bing-liang-ge-pai-xu-de-lian-biao-die-dai-di-gu/