题解

先排个序，保证连续，这样差就尽量的小，再用O(m-n) 判断最小值

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[100005];
int main(){
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= m; i++)  cin >> a[i];
	sort(a + 1, a + m + 1);
	int mx = 0x3f3f3f3f;
	for(int i = 1; i <= m - n + 1; i++){
		//cout << a[i + n - 1] << ' ' << a[i] << endl;
		mx = min(mx, a[i + n - 1] - a[i]);
	}cout << mx;

	return 0;
}

时间复杂度:$O(nlogn)$