21. Merge Two Sorted Lists【easy】

21. Merge Two Sorted Lists【easy】

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

解法一：

 /**
  * 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 || l2 == NULL) {
             return l1 ? l1 : l2;
         }

         ListNode * dummy = new ListNode(INT_MIN);
         ListNode * temp = dummy;

         while (l1 && l2) {
             if (l1->val > l2->val) {
                 dummy->next = l2;
                 l2 = l2->next;
             }
             else {
                 dummy->next = l1;
                 l1 = l1->next;
             }

             dummy = dummy->next;
         }

         if (l1 || l2) {
             dummy->next = l1 ? l1 : l2;
         }

         return temp->next;
     }
 };

由于最后是弄到list1中，但是我们不知道list1还是list2的第一个元素关系，最后结果的list1中的头结点可能会改变，所以需要引入dummy节点。

解法二：

 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;
         }
 }

参考了@yangliguang 的代码

解法三：

 public class Solution {
     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
         if (l1 == null) return l2;
         if (l2 == null) return l1;

         ListNode handler;
         if(l1.val < l2.val) {
             handler = l1;
             handler.next = mergeTwoLists(l1.next, l2);
         } else {
             handler = l2;
             handler.next = mergeTwoLists(l1, l2.next);
         }

         return handler;
     }
 }

参考了@RunRunCode 的代码

解法二和解法三都是递归，还没有完全弄明白……