<p>You are given an array of <code>k</code> linked-lists <code>lists</code>, each linked-list is sorted in ascending order.</p>

<p><em>Merge all the linked-lists into one sorted linked-list and return it.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> lists = [[1,4,5],[1,3,4],[2,6]]
<strong>Output:</strong> [1,1,2,3,4,4,5,6]
<strong>Explanation:</strong> The linked-lists are:
[
  1-&gt;4-&gt;5,
  1-&gt;3-&gt;4,
  2-&gt;6
]
merging them into one sorted list:
1-&gt;1-&gt;2-&gt;3-&gt;4-&gt;4-&gt;5-&gt;6
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> lists = []
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> lists = [[]]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>k == lists.length</code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= lists[i].length &lt;= 500</code></li>
	<li><code>-10<sup>4</sup> &lt;= lists[i][j] &lt;= 10<sup>4</sup></code></li>
	<li><code>lists[i]</code> is sorted in <strong>ascending order</strong>.</li>
	<li>The sum of <code>lists[i].length</code> will not exceed <code>10<sup>4</sup></code>.</li>
</ul>


/**
 * 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;

        // Priority queue to store the smallest element from all lists
        priority_queue<pair<int,
        ListNode*>,
        vector<pair<int,
        ListNode*>>,
        greater<pair<int,
        ListNode*>>>minHeap;

        // Insert the first element of each list into the minHeap
        for (ListNode* node : lists) {
            if (node) {
                minHeap.push({
                    node->val, node
                });
            node=node->next;
        }
    }

    // Build the merged list
    ListNode* head=new ListNode(0);
    ListNode* curr=head;

    while ( !minHeap.empty()) {
        auto[val,
        node]=minHeap.top();
        minHeap.pop();

        // Insert the value into the merged list
        curr->next=new ListNode(val);
        curr=curr->next;

        // Add the next element of the list to the minHeap if it exists
        if (node->next) {
            node=node->next;

            minHeap.push({
                node->val, node
            });
    }
}

return head->next;
}
}

;
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