leetcode 341. Flatten Nested List Iterator | 341. 扁平化嵌套列表迭代器（Java）

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题目

https://leetcode.com/problems/flatten-nested-list-iterator/

这标题，翻译的是人话吗？啥叫扁平化嵌套列表迭代器？。。

题解

比较像深度优先搜索。思路是：先 flatten 成 list 存好，需要的时候直接返回。感觉没有 get 到这个问题的精髓。评论区也有人质疑这种方法：

关于这个问题，我一开始也没打算先存成 list，仅存储调用的指针，没写出来，就先用了个 list 存数据了。

/**
 * // 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> {
    List<Integer> list;
    int index;

    public NestedIterator(List<NestedInteger> nestedList) {
        list = new ArrayList<>();
        flatten(list, nestedList);
        index = 0;
    }
    // Lying flat is standing up, horizontally. Lying flat is having a backbone.
    public void flatten(List<Integer> list, List<NestedInteger> nestedList) {
        for (NestedInteger next : nestedList) {
            if (next.isInteger()) {
                list.add(next.getInteger());
            } else {
                flatten(list, next.getList());
            }
        }
    }

    @Override
    public Integer next() {
        if (index < list.size()) return list.get(index++);
        else return null;
    }

    @Override
    public boolean hasNext() {
        return index < list.size();
    }
}

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

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