632. Smallest Range

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

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:
The given list may contain duplicates, so ascending order means >= here.
1 <= k <= 3500
-105 <= value of elements <= 105.

题目大意：
现在有已排好序的K个数组，问找到最小的一个范围[a,b]，使得每一个数组当中至少有一个元素落入这个区间。
思路：
这道题将会用到贪心的思想，首先假设k个数组满足一下条件
[a01,a02,a03,a04……]
[a11,a12,a13,a14……]
……
[ak1,ak2,ak3,ak4……]
假设存在a01

vector<int> smallestRange(vector<vector<int>>& nums) {        typedef vector<int>::iterator vi;        struct comp {            bool operator()(pair<vi, vi> p1, pair<vi, vi> p2) {                return *p1.first > *p2.first;            }        };        int lo = INT_MAX, hi = INT_MIN;        priority_queue<pair<vi, vi>, vector<pair<vi, vi>>, comp> pq;        for (auto &row:nums){            lo = min(lo,row[0]);            hi = max(hi,row[0]);            pq.push({row.begin(),row.end()});        }        vector<int> ans = {lo, hi};        while(true){            auto num =pq.top();            pq.pop();            num.first++;            if(num.first==num.second){                break;            }                pq.push(num);                lo = *pq.top().first;                if(*num.first>hi) hi =*num.first;                if(hi-lo<ans[1]-ans[0]){                    ans[0]=lo;                    ans[1]=hi;            }        }         return ans;    }