leetcode 632. Smallest Range 典型的移动窗口做法

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.

本题题意很简单，虽然是分开的k个list，但是可以合并到一起，然后做排序，使用移动窗口去做即可

本质上和本题leetcode 76. Minimum Window Substring 双指针 + Map + 移动窗口 是一样的做法

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution {public:    vector<int> smallestRange(vector<vector<int>>& nums)     {        vector<pair<int, int>> all;        int kind = nums.size();        for (int i = 0; i < kind; i++)        {            for (int a : nums[i])                all.push_back(make_pair(a, i));        }        sort(all.begin(), all.end());        map<int, int> mmp;        vector<int> res = {INT_MAX,INT_MAX};        int left = 0, count = 0 , minDiff = INT_MAX;        for (int right = 0; right < all.size(); right++)        {            mmp[all[right].second] += 1;            if(mmp[all[right].second] == 1)                count++;            while (count == kind && left <= right)            {                int diff = all[right].first - all[left].first;                if (diff < minDiff)                {                    minDiff = diff;                    res = { all[left].first , all[right].first};                }                else if (diff == minDiff && all[left].first < res[0])                    res = { all[left].first , all[right].first };                mmp[all[left].second] -= 1;                if (mmp[all[left].second] == 0)                    count--;                left++;            }        }        return res;    }};