跟踪的最低编号的数组中的 [英] Keep track of the lowest numbers in an array

问题描述

我想保持最低数量的分数的轨道，如果我找到那些球员的得分最低，我不希望他们在下一轮再次发挥。我已经得到了到存储这些低价值的球员到阵列的点，但我只希望他们能够暂时存储。

 的for（int i = 0; I＆LT; player.length;我++）{
  对于（INT J = 1; J＆LT; player.length; J ++）{
     如果（播放器[J]＆LT;播放器[I]）{
       分[I] = j的;
       的System.out.println（分[I] ++轮++ playerList.get（J））;
            }
        }
    }
 

解决方案

做2单独的回路来代替。其中找到最低的数字，第二个收集索引。

  INT minValue（最小值）= 1000000; //
的for（int i = 0; I＆LT; player.length;我++）{
  如果（播放器[1]  - ; minValue（最小值））{
    minValue（最小值）=玩家[I]
  }
}
INT J = 0;
的for（int i = 0; I＆LT; player.length;我++）{
  如果（播放器由[i] == minValue（最小值））{
    分[J] =我;
    J ++;
  }
}
 

I am trying to keep track of the the scores of the lowest numbers and if I find the lowest scores of those players I don't want them to play again in the next round. I have gotten to the point of storing those low player value into the array but I only want them to be stored ONCE.

 for(int i =0; i <  player.length; i++){
  for(int j =1; j <  player.length; j++){
     if(player[j] < player[i]){
       min[i] =j;
       System.out.println(min[i]+" "+round+" "+playerList.get(j));
            }
        }
    }

解决方案

Do 2 separate loops instead. One to find lowest number, second to collect indexes.

int minValue = 1000000; //
for(int i =0; i< player.length; i++){
  if(player[i] < minValue){
    minValue = player[i];
  }
}
int j =0;
for(int i =0; i< player.length; i++){
  if(player[i]==minValue){
    min[j]=i;
    j++;
  }
}

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