没什么可说的 直接暴力n3，当然也可以排序完之后把它优化一下这样更好。
if前两项的和小于第三项就直接结束第二重循环。 
import java.util.*;
import java.math.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
public class Main {
    public static void main(String args[])throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        in.nextToken();
        int n = (int) in.nval;
        long num[] = new long[n];
        for(int i=0;i<n;i++)
        {
            in.nextToken();
        num[i] = (long) in.nval;
        }
        boolean dui = false;
        long sum = 0;
        Arrays.sort(num);
        yi:for(int i=0;i<n-2;i++)
        {
           er:for(int k=i+1;k<n-1;k++)
            {
                for(int j=k+1;j<n;j++)
                {
                    if(num[i]+num[k]<num[j])
                    {
                        break er;
                    }
                    if(check(num[i],num[k],num[j])==true)
                       {
                           out.print(num[i]+" "+num[k]+" "+num[j]);
                           dui = true;;
                           break yi;
                       }
                    
                }
            }
        }
        if(dui==false)
            out.print("No solution");
        out.flush();
    }
    public static boolean check(long x,long y,long z)
    {
        if(x+y>z&&x+z>y&&y+z>x)
            return true;
        else
            return false;
    }
}