Title description:
Combine the two ascending linked lists into a new ascending linked list and return. The new linked list is composed by splicing all the nodes of the given two linked lists.
Example:
输入：1->2->4, 1->3->4 输出：1->1->2->3->4->4
Problem-solving ideas:
- Because the two linked lists are in order, we directly determine the value of a node in the two linked lists in a loop, which one is smaller and which one moves one bit backward, until the end of a linked list.
- Just follow the rest of the linked list that has not been judged to end.
Code:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* pp = new ListNode(0);
auto p = pp;
while(l1 && l2)
{
if(l1->val > l2->val){
p->next = l2;
p = p->next;
l2 = l2->next;
}
else{
p->next = l1;
p=p->next;
l1 = l1->next;
}
}
if(l1){ p->next = l1;}
else{ p->next = l2;}
pp = pp->next;
return pp;
}
};