多个数组内容拼接，每个数组分别根据索引位置，拼接内容

1，通过Hutool.ArrayUtil.zip方法进行处理（缺点，只能保留最短的数组内容）

public class Test {
    public static void main(String[] args) {
        String chinese = "张三；李四；王五";
        String english = "zhangsan；lisi";
        String age = "11；";

        String[] a1 = chinese.split("；");
        String[] a2 = english.split("；");
        String[] a3 = age.split("；");
        
        Map<String, String> zip = ArrayUtil.zip(a1, a2);
        System.out.println("zip = " + zip);
    }

}

结果

zip = {李四=lisi, 张三=zhangsan}

2，程序逻辑实现（短数组通过填充，解决数组越界）

public class Test {
    public static void main(String[] args) {
        String chinese = "张三；李四；王五";
        String english = "zhangsan；lisi";
        String age = "11；";

        String[] a1 = chinese.split("；");
        String[] a2 = english.split("；");
        String[] a3 = age.split("；");

        int[] lengthArr = {a1.length, a2.length, a3.length};
        int max = ArrayUtil.max(lengthArr);

        String[] t1 = transfer(max, a1);
        String[] t2 = transfer(max, a2);
        String[] t3 = transfer(max, a3);

        String[] todo = new String[max];

        for (int i = 0; i < max; i++) {
            todo[i] = t1[i] + " " + t2[i] + " " + t3[i];
        }
        System.out.println(Arrays.asList(todo));
    }

    private static String[] transfer(int max, String[] a1) {
        String[] m1 = new String[max];
        for (int i = 0; i < m1.length; i++) {
            if (a1.length - 1 >= i) {
                m1[i] = a1[i];
            } else {
                m1[i] = "";
            }
        }
        return m1;
    }
}

结果

[张三 zhangsan 11, 李四 lisi , 王五  ]