leetcode21

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
        //递归实现链表合并
            ListNode l = null;

            if (l1 == null && l2 != null)
            {
                return l2;
            }
            else if (l2 == null && l1 != null)
            {
                return l1;
            }
            else if (l1 == null && l2 == null)
            {
                return null;
            }
            else
            {
                if (l1.val < l2.val)
                {
                    if (l == null)
                    {
                        l = l1;
                    }
                    l.next = MergeTwoLists(l1.next, l2);
                }
                else
                {
                    if (l == null)
                    {
                        l = l2;
                    }
                    l.next = MergeTwoLists(l1, l2.next);
                }
                return l;
            }
    }
}

https://leetcode.com/problems/merge-two-sorted-lists/#/description

补充一个python的实现：

 1 class Solution:
 2     def mergeTwoLists(self, l1: 'ListNode', l2: 'ListNode') -> 'ListNode':
 3         if l1 == None:
 4             return l2
 5         if l2 == None:
 6             return l1
 7         if l1.val < l2.val:
 8             l1.next = self.mergeTwoLists(l1.next,l2)
 9             return l1
10         else:
11             l2.next = self.mergeTwoLists(l2.next,l1)
12             return l2

python非递归写法：

 1 class Solution:
 2     def mergeTwoLists(self, l1, l2):
 3         prehead = ListNode(-1)
 4         prev = prehead
 5         while l1 and l2:
 6             if l1.val <= l2.val:
 7                 prev.next = l1
 8                 l1 = l1.next
 9             else:
10                 prev.next = l2
11                 l2 = l2.next            
12             prev = prev.next
13 
14         # 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
15         prev.next = l1 if l1 is not None else l2
16 
17         return prehead.next

posted on 2017-04-20 11:17 Sempron2800+ 阅读(168) 评论(0) 编辑收藏举报