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.
递归解决很简单
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/**
* 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;
}
};
之前编译错误的代码附上求指点：
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/**
* 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* p,q,r,head;
if(l1=NULL && l2==NULL)
{
return NULL;
}
if(l1==NULL)
{
return l2;
}
if(l2==NULL)
{
return l1;
}
p=l1;
q=l2;
head=NULL;
if(p.val<=q.val)
{
head=p;
p=p->next;
}
else
{
head=q;
q=q->next;
}
r=head;
while(p!NULL && q!=NULL)
{
if(p->val<=q.val)
{
r->next=p;
r=p;
p=p->next;
}
else
{
r->next=q;
r=q;
q=q->next;
}
}
if(p==NULL)
{
while(q!=NULL)
{
r->next=q;
r=q;
q=q->next;
}
}
else
{
while(p!=NULL)
{
r->next=p;
r=p;
p=p->next;
}
}
return head;
}
};
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