23. Merge k Sorted Lists

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23. Merge k Sorted Lists

Description

Difficulty: Hard

Related Topics: Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort

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:

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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:

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Input: lists = []
Output: []

Example 3:

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Input: lists = [[]]
Output: []

Constraints:

  • k == lists.length
  • 0 <= k <= 104
  • 0 <= lists[i].length <= 500
  • -104 <= lists[i][j] <= 104
  • lists[i] is sorted in ascending order.
  • The sum of lists[i].length will not exceed 104.

Hints/Notes

Solution

Language: C++

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/**
 * 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) {}
 * };
 */
#include <queue>

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode *dummy = new ListNode();
        if (lists.size() == 0) {
            return dummy->next;
        }
        auto cmp = [](const ListNode* l1, const ListNode* l2) { return l1->val > l2->val; };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> pq(cmp);
        for(auto head: lists) {
            if (head) {
                pq.push(head);
            }
        }
        ListNode *p = dummy;
        while (pq.size() != 0) {
            ListNode *tmp = pq.top();
            p->next = tmp;
            p = p->next;
            pq.pop();
            if (tmp->next) {
                pq.push(tmp->next);
            }
        }
        return dummy->next;
    }
};