题目
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
合并两个有序链表。基本上就是按照归并排序的方法，再加上链表的操作就行了。
实现
/**
* 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 *ans, *pa;
if(l1==NULL) return l2;
if(l2==NULL) return l1;
if(l1->val < l2->val){
ans = l1;
l1 = l1->next;
}
else{
ans = l2;
l2 = l2->next;
}
pa=ans;
while(l1!=NULL && l2!=NULL){
if(l1->val < l2->val){
pa->next = l1;
l1=l1->next;
pa=pa->next;
}
else{
pa->next = l2;
l2=l2->next;
pa=pa->next;
}
}
while(l1!=NULL){
pa->next = l1;
l1=l1->next;
pa=pa->next;
}
while(l2!=NULL){
pa->next = l2;
l2=l2->next;
pa=pa->next;
}
return ans;
}
};
思考
重新理了下对于链表的理解。好久不用还是会生疏啊。
