Description
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.
Discussion
合并排序问题的最后一步。
算法的时间复杂度为O(n)。
C++ Code
/**
* 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 * answer = new ListNode(0);
ListNode * tmp = answer;
while(l1 && l2)
{
if(l1->val < l2->val)
{
tmp->next = l1;
l1 = l1->next;
}
else
{
tmp->next = l2;
l2 = l2->next;
}
tmp = tmp->next;
}
tmp->next = (l1? l1 : l2);
return answer->next;
}
};