java集合截取按照int数组里面的数值截取

List s = ["aa","sss","ddd","fff","ggg","hhhh"]
int[] a = [3,1,2]
//怎么把s按照数组a截取如下：
["aa","sss","ddd"]
["fff"]
["ggg","hhhh"]

public static void main(String[] args) {
String[] arrStr = {"aa","sss","ddd","fff","ggg","hhhh"};
//转换成list，可以通过下标删除元素
List list = Arrays.asList(arrStr);
/** 转换成ArrayList
* Arrays.asLisvt() 返回java.util.Arrays$ArrayList， 而不是ArrayList。
* Arrays$ArrayList和ArrayList都是继承AbstractList，remove，add等
* method在AbstractList中是默认throw UnsupportedOperationException而且不作任何操作。 ArrayList override这些method来对list进行操作，
* 但是Arrays$ArrayList没有override remove(int)，add(int)等，所以throw UnsupportedOperationException
* 所以需要转换成ArrayList
*/
List arrayList = new ArrayList(list);
//如果是list的话，就不需要以上的步骤了。
int[] a = {3,1,2};
for (int i = 0; i < a.length; i++) {
int num = a[i];
//初始化数组长度
String[] arr = new String[num] ;
//进行转成成数组,删除已经复制过的元素
for (int j = 0; j < num; j++) {
arr[j] = arrayList.get(0);
arrayList.remove(0);
}
System.out.println(Arrays.toString(arr));
}
}

list是这么用的吗？ 那个s 应该是个String数组才对吧？

List<List<String>>resultList=new ArrayList<List<String>>();
int start=0;
for(int len:a){
    List<String>res=new ArrayList<String>();
    for(int i=start;i<(start+len);i++){
        if(i<s.size()){
            res.add(s.get(i));
        }
    }
    resultList.add(res);
    start+=len;
}
//resList结果就是[["aa","sss","ddd"],["fff"],["ggg","hhhh"]]遍历就可以挨个获取

public class dddd {

public static void main(String[] args) {
    // TODO Auto-generated method stub
    //
String[] s = {"aa","sss","ddd","fff","ggg","hhhh"};
int[] a = {3,1,2};
int start=0;
int end=0;
for(int i=0;i<a.length;i++){
end=a[i]+start;
if(start==s.length){
    break;
}
String[] b=new String[a[i]];
for(int y=start;y<end;y++){
b[y-start]=s[y];    
}
show(b);
start=end;
}

}

private static void show(String[] b) {
    // TODO Auto-generated method stub
    String s ="[";
    for(int i=0;i<b.length;i++){
    if(i==0){
        s=s+b[i];
    }else{
        s=s+","+b[i];   
    }       
    }
    s+="]";
    System.out.println(s);
}

}

String []s = {"aa","sss","ddd","fff","ggg","hhhh"};
int[] a = {3,1,2};
int index = 0;
for (int i = 0; i < a.length; i++) {
String[] str = new String[a[i]];
for (int j = 0; j < str.length; j++) {
str[j] = s[j+index];
}
index+=a[i];
System.out.println(str);
}

    List<String> s = Arrays.asList("aa", "sss", "ddd", "fff", "ggg", "hhhh");
    int[] a = new int[]{3, 1, 2};

    List<List<String>> result = new ArrayList<>();
    int lastIdx = 0;
    for (int idx = 0; idx < a.length; idx++) {
        if (lastIdx >= s.size()) {
            break;
        }
        result.add(s.subList(lastIdx, Math.min(lastIdx + a[idx], s.size()))); // 考虑越界异常
        lastIdx += a[idx];
    }
    for (List<String> sl : result) {
        System.out.println(sl);
    }

亲测可以得到你想要的结果，麻烦采纳
public static void main(String[] args) {
String[] s = {"aa","sss","ddd","fff","ggg","hhhh"};
int[] a = {3,1,2};
int start = 0;
ArrayList list = new ArrayList();
for(int j=0;j<a.length;j++){
int len = a[j];
for(int i=start;i<(len+start);i++){
list.add(s[i]);
}
start = start + len;
System.out.println(list);
list.clear();
}
}

我的方法效率会比较高，时间复杂度O(n)

public static void main(String[] args) {

    String[] s = {"aa", "sss", "ddd", "fff", "ggg", "hhhh"};
    int[] a = {3, 1, 2};

    List<String> list = Arrays.asList(s);

    int intLen = a.length;

    int index = 0;
    for (int i = 0; i < intLen; i++) {
        System.out.println(list.subList(index,index + a[i]));;
        index = a[i];
    }
}