你如何找到一个整数二维数组中拥有最多0值的行？

huangapple
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英文:

How can i find the row which has the most 0 values in an int 2D array?

问题

如何找到具有最多0值的行在一个int二维数组中？

示例：

  1. int[][] arr = new int[4][5];
  2. arr[0][0] = 1;
  3. arr[0][1] = 2;
  4. arr[0][2] = 3;
  5. arr[0][3] = 4;
  7. arr[1][0] = 29;
  8. arr[1][1] = 28;
  9. arr[1][2] = 27;
  11. arr[2][0] = 3;
  12. arr[2][1] = 33;
  14. arr[3][0] = 41;
  15. arr[3][1] = 42;
  16. arr[3][2] = 43;
  17. arr[3][3] = 43;

答案应该是行索引2。

英文:

How can i find the row which has the most 0 values in an int 2D array?

Example:

  1. int[][] arr = new int[4][5]; 
  2. arr[0][0] = 1;
  3. arr[0][1] = 2;
  4. arr[0][2] = 3;
  5. arr[0][3] = 4;
  7. arr[1][0] = 29;
  8. arr[1][1] = 28;
  9. arr[1][2] = 27;
  11. arr[2][0] = 3;
  12. arr[2][1] = 33;
  14. arr[3][0] = 41;
  15. arr[3][1] = 42;
  16. arr[3][2] = 43;
  17. arr[3][3] = 43;

The answer is should be Row index 2

答案1

得分: 1

以下是翻译好的部分：

有几种方法可以实现你的目标。我会使用 stream 来完成这个任务：

  1. int[][] arr = new int[4][5]; 
  2. arr[0][0] = 1;
  3. arr[0][1] = 2;
  4. arr[0][2] = 3;
  5. arr[0][3] = 4;
  7. arr[1][0] = 29;
  8. arr[1][1] = 28;
  9. arr[1][2] = 27;
  11. arr[2][0] = 3;
  12. arr[2][1] = 33;
  14. arr[3][0] = 41;
  15. arr[3][1] = 42;
  16. arr[3][2] = 43;
  17. arr[3][3] = 43;
  19. IntStream.range(0, arr.length).boxed()
  20.         .map(row -> Map.entry(row, Arrays.stream(arr[row]).filter(i -> i != 0).count()))
  21.         .min(Comparator.comparing(Entry::getValue))
  22.         .map(Entry::getKey)
  23.         .ifPresent(row -> System.out.println("行: " + row));

输出

  1. 行: 2

更新

如果你想使用 for 循环，可以这样做：

  1. int row = -1;
  2. int maxZeros = 0;
  4. for (int i = 0; i < arr.length; i++) {
  5.     int zeros = 0;
  6.     for (int j = 0; j < arr[i].length; j++) {
  7.         if (arr[i][j] == 0) {
  8.             zeros++;
  9.         }
  10.     }
  11.     if (zeros > maxZeros) {
  12.         maxZeros = zeros;
  13.         row = i;
  14.     }
  15. }
  17. if (row > -1) {
  18.     System.out.println("行: " + row);
  19. }

输出

  1. 行: 2
英文:

There are several ways to achieve your goal. I would use streams for that job:

  1.     int[][] arr = new int[4][5]; 
  2.     arr[0][0] = 1;
  3.     arr[0][1] = 2;
  4.     arr[0][2] = 3;
  5.     arr[0][3] = 4;
  6.     
  7.     arr[1][0] = 29;
  8.     arr[1][1] = 28;
  9.     arr[1][2] = 27;
  10.     
  11.     arr[2][0] = 3;
  12.     arr[2][1] = 33;
  14.     arr[3][0] = 41;
  15.     arr[3][1] = 42;
  16.     arr[3][2] = 43;
  17.     arr[3][3] = 43;
  18.     
  19.     IntStream.range(0, arr.length).boxed()
  20.             .map(row -&gt; Map.entry(row, Arrays.stream(arr[row]).filter(i -&gt; i != 0).count()))
  21.             .min(Comparator.comparing(Entry::getValue))
  22.             .map(Entry::getKey)
  23.             .ifPresent(row -&gt; System.out.println(&quot;Row: &quot; + row));

Output

  1. Row: 2

Update

If you want to use a for loop you can do:

  1.     int row = -1;
  2.     int maxZeros = 0;
  3.     
  4.     for (int i = 0; i &lt; arr.length; i++) {
  5.         int zeros = 0;
  6.         for (int j = 0; j &lt; arr[i].length; j++) {
  7.             if (arr[i][j] == 0) {
  8.                 zeros++;
  9.             }
  10.         }
  11.         if (zeros &gt; maxZeros) {
  12.             maxZeros = zeros;
  13.             row = i;
  14.         }
  15.     }
  16.     
  17.     if (row &gt; -1) {
  18.         System.out.println(&quot;Row: &quot; + row);
  19.     }

Output

  1. Row: 2

huangapple
  • 本文由 发表于 2020年7月23日 07:32:41
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  • java
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