[C++] Leetcode 23. Merge k Sorted Lists (Hint, Solution)

Problem

https://leetcode.com/problems/merge-k-sorted-lists

Merge k Sorted Lists - LeetCode

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 <= 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.

 

 

 

 

Hint

Try using min heap structure.

 

 

Solution

 

 

 

Code

Simple

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<int, vector<int>, greater<int>> minHeap;

        for (auto list : lists)
            for (ListNode* it=list; it != nullptr; it = it->next)
                minHeap.push(it->val);

        ListNode* head = nullptr;

        for (ListNode* current = nullptr; !minHeap.empty();) {
            ListNode* node = new ListNode(minHeap.top());
            if (head == nullptr)
                head = node;
            else
                current->next = node;
            current = node;
            minHeap.pop();
        }

        return head;
    }
};

 

Whole Code with Test Case

// https://leetcode.com/problems/merge-k-sorted-lists
#include <iostream>
#include <vector>
#include <queue>

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) {
        priority_queue<int, vector<int>, greater<int>> minHeap;

        for (auto list : lists)
            for (ListNode* it=list; it != nullptr; it = it->next)
                minHeap.push(it->val);

        ListNode* head = nullptr;

        for (ListNode* current = nullptr; !minHeap.empty();) {
            ListNode* node = new ListNode(minHeap.top());
            if (head == nullptr)
                head = node;
            else
                current->next = node;
            current = node;
            minHeap.pop();
        }

        return head;
    }
};

int main() {
    Solution s = Solution();
}

 

 

 

Complexity

Time

Time complexity will be O(n*log n).
First loop has a time complexity of O(n* log n), and the second loop also has a time complexity of O(n*log n).

 

Space

Space complexity will be O(n).

 

 

 

Best Solution (Pinned)

 

 

 

Overall