Implement an iterator to flatten a 2d vector.
For example, Given 2d vector =
[
[1,2],
[3],
[4,5,6]
] By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,2,3,4,5,6].
分析：https://segmentfault.com/a/1190000003791233
用一个数组表示每个List的迭代器，然后再记录一个变量，用来表示当前用到了第几个迭代器。
public class Vector2D {
List<Iterator<Integer>> its;
int curr = ;
public Vector2D(List<List<Integer>> vec2d) {
this.its = new ArrayList<Iterator<Integer>>();
for (List<Integer> l : vec2d) {
// 只将非空的迭代器加入数组
if (l.size() > ) {
this.its.add(l.iterator());
}
}
}
public int next() {
Integer res = its.get(curr).next();
// 如果该迭代器用完了，换到下一个
if (!its.get(curr).hasNext()) {
curr++;
}
return res;
}
public boolean hasNext() {
return curr < its.size() && its.get(curr).hasNext();
}
}