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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.
分析：思路比较简单，遍历两个有序链表，每次指向最小值。
code如下：
/**
* 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) {
ListNode *h = new ListNode(0), *p = h;
while(l1 != NULL && l2 != NULL)
{
if(l1 -> val < l2 -> val)
{
p -> next = l1;
l1 = l1 -> next;
}
else
{
p -> next = l2;
l2 = l2 -> next;
}
p = p -> next;
}
if(l1 != NULL)
p -> next = l1;
if(l2 != NULL)
p -> next = l2;
return h -> next;
}
};
leetcode:Merge Two Sorted Lists（有序链表的归并）
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原文地址：http://www.cnblogs.com/carsonzhu/p/4857657.html
