Given array of k sorted linked-lists, merge all into 1 sorted list
Ex. lists = [[1,4,5],[1,3,4],[2,6]] -> [1,1,2,3,4,4,5,6]
Min heap -> optimize space w/ divide-and-conquer, merge 2 each time
Time: O(n log k)
Space: O(n) -> O(1)
*/
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
int n = lists.size();
if (n == 0) {
return NULL;
}
while (n > 1) {
for (int i = 0; i < n / 2; i++) {
lists[i] = mergeTwoLists(lists[i], lists[n - i - 1]);
}
n = (n + 1) / 2;
}
return lists.front();
}
private:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
if (list1 == NULL && list2 == NULL) {
return NULL;
}
if (list1 == NULL) {
return list2;
}
if (list2 == NULL) {
return list1;
}
ListNode* head = NULL;
if (list1->val <= list2->val) {
head = list1;
list1 = list1->next;
} else {
head = list2;
list2 = list2->next;
}
ListNode* curr = head;
while (list1 != NULL && list2 != NULL) {
if (list1->val <= list2->val) {
curr->next = list1;
list1 = list1->next;
} else {
curr->next = list2;
list2 = list2->next;
}
curr = curr->next;
}
if (list1 == NULL) {
curr->next = list2;
} else {
curr->next = list1;
}
return head;
}
};