NetEase 2016 Internship Research Engineer 3 programming Questions-1

Xiao Ming accompanied Xiao Red to see the diamonds, they from a bunch of diamonds randomly extracted two and compare their weight. The weight of these diamonds varies. After they had been comparing for some time, they had a fancy for two diamonds G1 and G2. Now, please judge which of the two diamonds is heavier according to the information you have previously compared.

Given the number of two diamonds g1,g2, the number starts at 1, given the relationship array vector, where the elements are some binary groups, the first element is the number of the heavier diamond in a comparison, and the second element is the number of the lighter diamond. The last number of comparisons before given N. Please return the relationship between the two diamonds, if G1 more return 1,g2 more return-1, can not be judged to return 0. The input data is guaranteed to be legal and there will be no contradiction.

Test examples:

2,3,[[1,2],[2,4],[1,3],[4,3]],4

Returns: 1

import java.util.*;p ublic class cmp {    public static void  main (String[] args)  {          // todo  auto-generated method stub          int[][]  records = { { 1, 2 }, { 2, 4 }, { 1, 3  }, { 4, 3 } };           SYSTEM.OUT.PRINTLN (CMP (2, 3, records, 4));      }   &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;PUBLIC&NBSP;STATIC&NBSP;INT&NBSP;CMP (Int g1, int  g2, int[][] records, int n)  {        //  write code here        hashset<integer> max  = new hashsEt<> ();          hashset<integer> min  = new hashset<> ();          for  (int  i = 0; i < n; i++)  {               if  (RECORDS[I][0]&NBSP;==&NBSP;G1)  {                   min.add (records[i][1 ]);              }               if  (RECORDS[I][1]&NBSP;==&NBSP;G1)  {                   max.add (records[i][0]);               }    &nbSp;     }            for   (int i = 0; i < n; i++)  {               if  (records[i][0] != g1 &&  RECORDS[I][1]&NBSP;!=&NBSP;G1)  {                   if  (Max.contains (records[i][1))  {                       max.add ( Records[i][0]);                   }                   if  (Min.contains (records[i][0))  {            &nBsp;          min.add (records[i][1]);                   }               }           }            if  ( Max.contains (G2)  && !min.contains (G2))  {               return -1;           } else if  (!max.contains (G2)  && min.contains (G2))  {               return 1;           } else {               return 0;      }   }} 

NetEase 2016 Internship Research Engineer 3 programming Questions-1