<p>You have <code>k</code> lists of sorted integers in <strong>non-decreasing&nbsp;order</strong>. Find the <b>smallest</b> range that includes at least one number from each of the <code>k</code> lists.</p>

<p>We define the range <code>[a, b]</code> is smaller than range <code>[c, d]</code> if <code>b - a &lt; d - c</code> <strong>or</strong> <code>a &lt; c</code> if <code>b - a == d - c</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]
<strong>Output:</strong> [20,24]
<strong>Explanation: </strong>
List 1: [4, 10, 15, 24,26], 24 is in range [20,24].
List 2: [0, 9, 12, 20], 20 is in range [20,24].
List 3: [5, 18, 22, 30], 22 is in range [20,24].
</pre>

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

<pre>
<strong>Input:</strong> nums = [[1,2,3],[1,2,3],[1,2,3]]
<strong>Output:</strong> [1,1]
</pre>

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

<ul>
	<li><code>nums.length == k</code></li>
	<li><code>1 &lt;= k &lt;= 3500</code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 50</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code>&nbsp;is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        int n = nums.size();
        // [-value, idNums, index], the choose num in every list
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> que;

        int left = INT_MAX;
        int right = INT_MIN;
        for(int i = 0; i < n; i++){
            left = min(left, nums[i][0]);
            right = max(right, nums[i][0]);
            que.push({nums[i][0], i, 0});
        }
        int curMin = left;
        int curMax = right;
        while(true){
            int value = que.top()[0];
            int idNums = que.top()[1];
            int index = que.top()[2];
            que.pop();
            // break the loop if any of the list reach the end.
            if(index == nums[idNums].size() - 1){
                break;
            }
            // update the answer.
            int nValue = nums[idNums][index + 1];
            curMax = max(curMax, nValue);
            curMin = min(nValue, que.top()[0]);
            if(curMax - curMin < right - left || (curMax - curMin == right - left && curMax < left)){
                right = curMax;
                left = curMin;
            }

            // push the next into the queue.
            que.push({nValue, idNums, index + 1});
        }
        return {left, right};
    }
};
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