java - 获取一个数组，判断该数字是否更接近右/左零

Question

我正在尝试编写一个获取整数数组的代码，检查每个索引的值，如果索引的值为零，则继续下去，直到找到值不为零的索引，然后根据情况更改其值数组中越接近零。例如:

  BEFORE  ( 0, 1 , 1 , 1 , 1 , 0 , 1 , 1 , 1 , 1 , 1 , 0) 

  AFTER ( 0, 1 , 2 , 2 , 1 , 0 , 1 , 2 , 3 , 2 , 1 , 0) 

这是我的代码:

       public static void zeroDistance (int [] a)
   {
       int counterToZero = 0;
       int counterToZeroII = 0;
       for ( int i = 0; i < a.length; i++ ){
                counterToZero = 0;
                counterToZeroII = -1;
                if ( a[i] != 0 ){
                  for ( int j = i; a[j] != 0; j++)
                  { // From i to the next zero
                          counterToZero++;
                  }
                  for ( int h = i; a[h] != 0 ; h--) // From i to first zero (goes beckwards)
                  {
                        counterToZeroII++;
                  }
                  if ( counterToZero > counterToZeroII ) 
                  a[i] = counterToZero;
                  else if (counterToZero < counterToZeroII)
                  a[i] = counterToZeroII;
                  else if ( counterToZero == counterToZeroII)
                  a[i] = counterToZero;
                }
       }
       int[] b = a;
       for (int h = 0; h < b.length; h++)
        System.out.println(b[h] + "/n");
   }

这是我得到的输出，我有点不知道问题出在哪里，我仍在查看代码并试图检测我做错了什么，如果你能帮助我并指出问题所在我可能错了，这会很有帮助。另外，如果你知道另一种“排序检查”的方法，比如使用递归，我会非常感激，我正在尝试习惯使用递归，但对我来说仍然很难实现，因此我在时刻。

0/n 2/n 1/n 0/n 5/n 4/n 3/n 3/n 4/n 0/n

Answer 1

我可以向您推荐一种简单的方法。首先从左到右，只考虑最左边的零。然后从右到左，只考虑最接近的右边零。如果之前的距离较小，则将其替换为新的距离。

以下是它的实现方式:

public static void zeroDistance (int [] a) {
    // closest == -1 means no zero was found yet
    int closest = -1;
    for (int i=0 ; i<a.length ; i++) 
       if (a[i] == 0) 
           closest = 0;
       else {         
           // short version
           // a[i] = closest == -1 ? Integer.MAX_VALUE : ++closest;

           // simpler version for your to understand
           if (closest == -1) 
               a[i] = Integer.MAX_VALUE;
           else {
               closest++;
               a[i] = closest;
           }                    
       }
    closest = -1;
    for (int i=a.length-1 ; i>=0 ; i--) 
       if (a[i] == 0)                            
           closest = 0;
       else if (closest != -1 && a[i] > ++closest) 
           a[i] = closest;       
}

示例的输出:

0, 1, 2, 2, 1, 0, 1, 2, 3, 2, 1, 0