Hard
You are given an array of k
linked-lists lists
, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
Example 1:
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted list: 1->1->2->3->4->4->5->6
Example 2:
Input: lists = []
Output: []
Example 3:
Input: lists = [[]]
Output: []
Constraints:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i]
is sorted in ascending order.lists[i].length
won’t exceed 10^4
.#include <vector>
using namespace std;
/**
* 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) {
if (lists.empty()) {
return nullptr;
}
return mergeKLists(lists, 0, lists.size());
}
private:
ListNode* mergeKLists(vector<ListNode*>& lists, int leftIndex, int rightIndex) {
if (rightIndex > leftIndex + 1) {
int mid = (leftIndex + rightIndex) / 2;
ListNode* left = mergeKLists(lists, leftIndex, mid);
ListNode* right = mergeKLists(lists, mid, rightIndex);
return mergeTwoLists(left, right);
} else {
return lists[leftIndex];
}
}
ListNode* mergeTwoLists(ListNode* left, ListNode* right) {
if (left == nullptr) {
return right;
}
if (right == nullptr) {
return left;
}
ListNode* res;
if (left->val <= right->val) {
res = left;
left = left->next;
} else {
res = right;
right = right->next;
}
ListNode* node = res;
while (left != nullptr || right != nullptr) {
if (left == nullptr) {
node->next = right;
right = right->next;
} else if (right == nullptr) {
node->next = left;
left = left->next;
} else {
if (left->val <= right->val) {
node->next = left;
left = left->next;
} else {
node->next = right;
right = right->next;
}
}
node = node->next;
}
return res;
}
};