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.
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表示有点难，想了很久。
递归的算法：
1,3,10,12
2,3,4,7,9,11
递归的算法就是，首先，首节点1和2进行比较，1比2小，所以3再和2比较，因为2比3小，所以2被加入，3和3比较。（我之前想错了，不应该是两两比较，应该是较大的和较小的剩下的比较）
/** * 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; if (l2 == NULL) return l1; ListNode *ret = NULL; if (l1->val < l2->val) { ret = l1; ret->next = mergeTwoLists(l1->next, l2); } else { ret = l2; ret->next = mergeTwoLists(l1, l2->next); } return ret; } };
leetcode Merge Two Sorted Lists
原文：http://www.cnblogs.com/LUO77/p/4986161.html