AcWing 36. 合并两个排序的链表
原题链接
简单
作者:
daniellee
,
2019-04-10 21:12:23
,
所有人可见
,
阅读 838
C++ 代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* merge(ListNode* l1, ListNode* l2) {
ListNode* h = new ListNode(0);
ListNode* h1 = h;
while((l1 != NULL && l2 != NULL) || (l1!=NULL && l2==NULL) || (l2!=NULL && l1==NULL) ){
// cout << l1<<" "<<l2<<endl;
if (l1!=NULL && l2!=NULL){
if (l1->val < l2->val){
h -> next = new ListNode(l1->val);
h = h->next;
l1=l1->next;
}
else{
h -> next = new ListNode(l2->val);
h = h->next;
l2 = l2->next;
}
}
if (l1==NULL && l2!=NULL){
h -> next = new ListNode(l2->val);
h = h->next;
l2 = l2->next;
}
else if(l2==NULL && l1!=NULL){
h -> next = new ListNode(l1->val);
h = h->next;
l1=l1->next;
}
}
return h1->next;
}
};