java如何在数组中迭代并跳过一些？

Question

例如，我的Java程序中有一个数组：

String nums[] = {"a", "b", "c", "d", "e", "f", "g", "h" ...}

我想写for循环，循环遍历数组，每2个和3个字母，将它们存储在数组中的两个连续索引中，跳过第4个，取5个和6个字母，将它们存储在数组中的两个连续索引中，跳过第7个，并继续为未知大小的数组执行此操作

所以最后一个数组是nums2 = {"b", "c", "e", "f", "h", "i"...}

Answer 1

请永远分享你迄今为止的尝试。人们会更愿意帮助你。否则，你最应该得到的就是伪代码。试试这样：

    for (1 to length)
    {
        if( i % 3 != 0)
        add to new array
    }

Answer 2

它运行并打印out = b, c, e, f, h, i

public class Skip {
    public static String[] transform(String[] in) {
        int shortenLength = (in.length / 3) + ((in.length % 3 > 0) ? 1 : 0);
        int newLength = in.length - shortenLength;
        String[] out = new String[newLength];
        int outIndex = 0;
        for (int i = 0; i < in.length; i++) {
            if (i % 3 != 0) {
                out[outIndex++] = in[i];
            }
        }
        return out;
    }

    public static void main(String[] args) {
        String[] nums = {"a", "b", "c", "d", "e", "f", "g", "h", "i" };
        String[] out = transform(nums);
        System.out.println("out = " + String.join(", ", out));
    }
}

Answer 3

 String[] array = {"a", "b", "c", "d", "e", "f", "g", "h" ...} //Consider any datatype 
 for(int i =1; i<array.length;i++) {
 if(i%3 == 0) {
 }
 else {
 System.out.println(a[array]);
 }

}

这样，它将跳过第4个元素、第7个元素、第10个元素、第13个元素，这些元素对应的索引值是3的倍数，我们将根据if条件跳过这些索引元素

Answer 4

for(int num : nums){
    if(num % 3 == 1) Continue;
    System.out.print(num + " ");
}

如上所述的示例java代码

Answer 5

可以在for循环中使用if语句，该语句将从数组中的第二项开始，每三个字母跳过一次

int j=0;    //separate counter to add letters to nums2 array
for(int i=0; i<nums.length; i++) {    //starts from 1 to skip the 0 index letter
    if (i%3 != 0) {    //should include every letter except every third
        nums2[j] = nums[i];
        j++;
    }
}

Answer 6

要获得最简洁的方式，请使用Java 9 streams：

String[] nums2 = IntStream.range(0, nums.length)
    .filter(i -> i % 3 != 0)
    .mapToObj(i -> nums[i])
    .toArray(String[]::new);