原题目：https://leetcode-cn.com/problems/smallest-range-covering-elements-from-k-lists/

 

思路：

采用hash、滑动窗口和双指针的方法，当窗口不包含每个数组中的元素时，右指针右移，直到窗口包含。包含以后，左指针右移，直到不包含，此时记录betsleft,bestright

 

代码：

class Solution {
public:
    vector<int> smallestRange(vector<vector<int>>& nums) {
        
        int n = nums.size();
        unordered_map<int, vector<int>> indices;
        int xMin = INT_MAX, xMax = INT_MIN;
        for (int i = 0; i < n; ++i) {
            for (const int& x: nums[i]) {
                indices[x].push_back(i);
                xMin = min(xMin, x);
                xMax = max(xMax, x);
            }
        }

        vector<int> freq(n);
        int inside = 0;
        int left = xMin, right = xMin - 1;
        int bestLeft = xMin, bestRight = xMax;

        while (right < xMax) {
            ++right;
            if (indices.count(right)) {
                for (const int& x: indices[right]) {
                    ++freq[x];
                    if (freq[x] == 1) {
                        ++inside;
                    }
                }
                while (inside == n) {
                    if (right - left < bestRight - bestLeft) {
                        bestLeft = left;
                        bestRight = right;
                    }
                    if (indices.count(left)) {
                        for (const int& x: indices[left]) {
                            --freq[x];
                            if (freq[x] == 0) {
                                --inside;
                            }
                        }
                    }
                    ++left;
                }
            }
        }

        return {bestLeft, bestRight};
    }
};