方法一：
/**
* 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) return l2;
if(l2 == NULL) return l1;
ListNode* head;
ListNode* end;
head=(ListNode*)malloc(sizeof(ListNode));
head->next=NULL;
end=head;
while(l1&&l2)
{
if(l1->val<=l2->val)
{
end->next=l1;
end=end->next;
l1=l1->next;
}
else
{
end->next=l2;
end=end->next;
l2=l2->next;
}
}
if(l1)
{
end->next=l1;
}
if(l2)
{
end->next=l2;
}
return head->next;
}
};
方法二：
/////来自网上的方法
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
ListNode *l, *ans;
if(l1->val > l2->val) {
l = l1;
l1 = l2;
l2 = l;
}
ans = l1;
while(l2 != NULL) {
if(l1 == NULL) return ans;
if(l1->val <= l2->val) {
if(l1->next == NULL||l1->next->val >= l2->val){
l = l2;
l2 = l2->next;
l->next = l1->next;
l1->next = l;
}
l1 = l1->next;
}
else {
l2 = l2->next;
}
}
return ans;