您可以在 range 矩阵的行索引，以及 filter 把那些包含 anyMatch 的 1 :

int[][] runners = {
        {21, 14, 1, 21, 19, 5, 2},
        {4, 27, 1, 7, 25, 19, 19},
        {12, 20, 25, 11, 30, 14, 15},
        {5, 8, 5, 7, 1, 2, 13},
        {20, 2, 14, 4, 25, 1, 1},
        {14, 20, 20, 15, 11, 18, 16}};

int[] winners = IntStream
        // iterate over indices of
        // the rows of 2d array
        .range(0, runners.length)
        // filer rows where at least
        // one element matches '1'
        .filter(i -> Arrays.stream(runners[i])
                .anyMatch(j -> j == 1))
        // array of indices
        .toArray();

System.out.println(Arrays.toString(winners)); // [0, 1, 3, 4]

要跟踪在比赛中已经获得第一名的跑步者，可以使用如下布尔数组

boolean[] firstPlaceRunner = new boolean[A[0].length];

上述布尔数组的第i个位置存储以下信息：
如果第i名选手在任何一场比赛中获得第一名。
错误-如果第i名选手无法在任何比赛中获得第一名。
因此，现在在您的代码中，您只能在当前运行程序尚未获得第一名时递增计数器。您的代码如下所示：

public static void main(String[] args) {
    int[][] A = {{21, 14, 1, 21, 19, 5, 2},
            {4, 27, 1, 7, 25, 19, 19},
            {12, 20, 25, 11, 30, 14, 15},
            {5, 8, 5, 7, 1, 2, 13},
            {20, 2, 14, 4, 25, 1, 1},
            {14, 20, 20, 15, 11, 18, 16}};
    int counter = 0;
    // A[0].length gives the number of runner in any race
    boolean[] firstPlaceRunner = new boolean[A[0].length];
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 7; j++) {
            if (A[i][j] == 1) {
                if (!firstPlaceRunner[j]) {
                    counter++;
                    // because now jth runner has secured first
                    // place, and in future we don't want to
                    firstPlaceRunner[j] = true;
                    // double count it
                }
            }
        }
    }
    System.out.println("result:");
    System.out.print(counter);
}

注意：作为对我们算法的改进，您可以 break 因为只能有一个第一名优胜者，所以在代码中的下一行后面的语句：

firstPlaceRunner[j] = true;
break;

所以基本上，你是说，如果我们已经找到了一个给定比赛的第一名，那么就不要检查其他选手。

在if语句中添加中断符。

public class MyClass {
public static void main(String[] args) {
    int A[][] = {{21, 14, 1, 21, 19, 5, 2},
            {4, 27, 1, 7, 25, 19, 19 },
            {12, 20, 25, 11, 30, 14, 15},
            {5, 8, 5, 7, 1, 2, 13},
            {20, 2, 14, 4, 25, 1, 1},
            {14, 20, 20, 15, 11, 18, 16}};
    int i, j;
    int counter = 0;

    System.out.println("result:");
    for (i=0; i<6; i++) {
        for (j=0; j<7; j++) {
            if (A[i][j] == 1) {
                counter++;
                break;
            }
        }

        }
    System.out.print(counter);
    }
}