本文介绍了首先按列填充锯齿状的二维数组的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想编写一个函数,它接受一个二维数组并用 1...n
填充它,但首先计算列而不是行:
I want to write a function that takes an 2d array and fills it with 1...n
but counting the columns first instead of the rows:
input = {{0, 0, 0, 0}, {0}, {0}, {0, 0}};
the output should be: {{1, 5, 7, 8}, {2}, {3}, {4, 6}};
如果我要循环遍历行然后得到列:
if i were to loop through rows and then colums i get:
private static void fill1(int[][] input) {
int count = 1;
for (int i = 0; i < input.length; i++) {
for (int j = 0; j < input[i].length; j++) {
input[i][j] = count;
count++;
}
}
}
如何首先遍历列?
推荐答案
要首先按列填充二维数组,您可以使用两个嵌套的流.在锯齿状二维数组的情况下,当您事先不知道每行中的列数时,您可以在列仍然存在的情况下遍历外部流.
To populate a 2d array first by columns, you can use two nested streams. In case of a jagged 2d array, when you don't know beforehand the number of the columns in each row, in an outer stream you can traverse while the columns are still present.
/**
* @param arr array that should be populated.
* @return maximum row length, i.e. columns count.
*/
private static long populate(int[][] arr) {
AtomicInteger counter = new AtomicInteger(1);
return IntStream
// traverse through the array columns
.iterate(0, i -> i + 1)
// process the array rows where
// this column is present
.mapToLong(i -> Arrays.stream(arr)
// filter those rows where
// this column is present
.filter(row -> row.length > i)
// assign a value to the element and increase the counter
.peek(row -> row[i] = counter.getAndIncrement())
// count of rows where this column is present
.count())
// while the columns are still present
.takeWhile(i -> i > 0)
// max columns count
.count();
}
public static void main(String[] args) {
int[][] arr = {{0, 0, 0, 0, 0, 0}, {0, 0}, {0}, {0, 0, 0}};
System.out.println("Max columns count: " + populate(arr));
System.out.println(Arrays.deepToString(arr));
}
输出:
Max columns count: 6
[[1, 5, 8, 10, 11, 12], [2, 6], [3], [4, 7, 9]]
另见:如何以循环方式合并 3 个 ArrayList 来创建新列表?上>
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