Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
可以借鉴Merge two sorted lists的方法,在他的基础上用分治算法。
/**
* 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* merge = new ListNode(0);
ListNode* head = merge;
while(l1&&l2)
{
if(l1->val < l2->val)
{
merge->next = l1;
l1 = l1->next;
merge = merge->next;
}
else
{
merge->next = l2;
l2 = l2->next;
merge = merge->next;
}
}
if(l1)
{
merge->next = l1;
}
if(l2)
{
merge->next = l2;
}
return head->next;
}
ListNode* mergeKLists(vector<ListNode*>& lists) {
int n = lists.size();
if(n == 0) return NULL;
while(n>1)
{
int k = (n+1)/2;
for(int i = 0; i < n/2; i++)
{
lists[i] = mergeTwoLists(lists[i], lists[i+k]);
}
n = k;
}
return lists[0];
}
};