<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* merge(ListNode* a, ListNode* b) {
       ListNode*c=new ListNode(100);
       ListNode*temp=c;
       while(a!=NULL && b!=NULL){
        if(a->val<=b->val){
            temp->next=a;
            temp=a;
            a=a->next;
        }
        else{
            ListNode*t=new ListNode(b->val);
            temp->next=b;
            temp=b;
            b=b->next;
        }
       }
       if(a==NULL) temp->next=b;
       else temp->next=a;
       return c->next;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if(lists.size()==0) return NULL;
        while(lists.size()>1){
            ListNode*a=lists[0];
            lists.erase(lists.begin());
            ListNode*b=lists[0];
            lists.erase(lists.begin());
            ListNode*temp=merge(a,b);
            lists.push_back(temp);
        }
        return lists[0];
    }
};
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