腾讯//合并两个有序链表

将两个有序链表合并为一个新的有序链表并返回。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：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

/**
 * 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 temp1 = l1;
        ListNode temp2 = l2;
        
        ListNode ret = new ListNode(0);
        
        ListNode ret1 = ret;
        while(temp1!=null&&temp2!=null){
            if(temp1.val>temp2.val){
                ret.next = temp2;
                ret = ret.next;
                temp2 = temp2.next;
            }else{
                ret.next = temp1;
                ret = ret.next;
                temp1 = temp1.next;
            }
        }
        if(temp1==null&&temp2!=null)
            ret.next = temp2;
        if(temp2==null&&temp1!=null)
            ret.next = temp1;
        return ret1.next;
    }
}

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) {
        ListNode temp1 = l1;
        ListNode temp2 = l2;
        ListNode ret = getResult(temp1, temp2);
        return ret;
    }
    public ListNode getResult(ListNode t1, ListNode t2){
        ListNode t0 = new ListNode(0);
        ListNode head = t0;
        if(t1 == null) return t2;
        if(t2 == null) return t1;
        if(t1.val>t2.val){
            t0.next = t2;
            t0 = t0.next;
            t0.next = getResult(t1, t2.next);
        }else{
            t0.next = t1;
            t0 = t0.next;
            t0.next = getResult(t1.next,t2);
        }
        return head.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL)
            return l2;
        else if(l2 == NULL)
            return l1;
        
        ListNode* pMergeHead = NULL;
        if(l1->val < l2->val){
            pMergeHead = l1;
            pMergeHead->next = mergeTwoLists(l1->next, l2);
        }else{
            pMergeHead = l2;
            pMergeHead->next = mergeTwoLists(l1, l2->next);
        }
        return pMergeHead;
    }
};