632. Smallest Range

问题描述：

You have k lists of sorted integers in ascending order. Find the smallest range that includes at least one number from each of the klists.

We define the range [a,b] is smaller than range [c,d] if b-a < d-c or a < c if b-a == d-c.

Example 1:

Input:[[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
Output: [20,24]
Explanation: 
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].

Note:

1. The given list may contain duplicates, so ascending order means >= here.
2. 1 <= k <= 3500
3. -105 <= value of elements <= 105.
4. For Java users, please note that the input type has been changed to List<List<Integer>>. And after you reset the code template, you'll see this point.

解题思路：

首先我们将所有数组混合到一起，为了标识它原本属于哪个数组，我们用pair来存储，pair.first为数值， pair.second为它来自哪个数组。

然后我们用sort将其排序。

排序后可以用滑动窗口来解决这个问题。

即，找一个窗口两端差最小的窗口，窗口里的值必须每个链表中的至少一个值。

可以参考 76. Minimum Window Substring

不过将比较值设为v[i].first - v[l].first。

同时我们也用一个map来存储k个链表中的值的个数。

代码：

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        unordered_map<int, int> m;
        vector<pair<int,int>> v;
        vector<int> ret;
        for(int i = 0; i < nums.size(); i++){
            for(int j = 0; j < nums[i].size(); j++){
                v.push_back(pair<int,int>(nums[i][j], i));
            }
            m[i] = 1;
        }
        sort(v.begin(), v.end());
        int count = 0;
        int l = 0, len = INT_MAX;
        int k = nums.size();
        for(int i = 0; i < v.size(); i++){
            pair<int, int> p = v[i];
            m[p.second]--;
            if(m[p.second] > -1)
                count++;
            while(count == k && l <= i){
                if(len > (v[i].first - v[l].first)){
                    len = v[i].first - v[l].first;
                    ret = {v[l].first, v[i].first};
                }
                if(++m[v[l].second] == 1)
                    count--;
                l++;
            }
        }
        return ret;
    }
};