题目链接： 力扣

 迭代器特点：惰性，如果你要一个结果，我就算一个（或是一小部分）结果出来，而不是一次把所有结果都算出来。

故解法如下：

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return empty list if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class NestedIterator implements Iterator<Integer> {
    LinkedList<NestedInteger> list ;
    
    public NestedIterator(List<NestedInteger> nestedList) {
       list = new LinkedList<>(nestedList);
        
    }

    @Override
    public Integer next() {
        return list.remove(0).getInteger();
    }

    @Override
    public boolean hasNext() {
        while(!list.isEmpty()&&!list.get(0).isInteger()){
            List<NestedInteger> list1= list.remove(0).getList();
            for(int i=list1.size()-1;i>=0;i--){
                list.addFirst(list1.get(i));
            }
        }
        return !list.isEmpty();
    }
}

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i = new NestedIterator(nestedList);
 * while (i.hasNext()) v[f()] = i.next();
 */

注意：

1. list长度使用size()而不是length()

2. LinkedList是双向链表，我们可以使用它进行头部的添加与删除，注意声明的时候不能使用List声明，因为list类无addFirst方法

3.  list的构造函数可以传入另一个list参数。

4. isEmpty方法判断list是否为空。