扁平化嵌套列表迭代器

LeetCode每日一题，341. Flatten Nested List Iterator

先看题目描述

大意就是给定一个多维列表，实现一个迭代器可以扁平化的迭代它

算法和思路

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把多维列表转化为一维列表即可，对于列表里的整数，就将其直接加到一维列表中，对于列表里嵌套的列表，就使用 dfs 来处理，最后使用一维列表自带的迭代器进行迭代即可

算法源码

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/**
 * // 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 null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class NestedIterator implements Iterator<Integer> {
    private List<Integer> list = new ArrayList<>();
    private Iterator<Integer> iterator;

    public NestedIterator(List<NestedInteger> nestedList) {
        dfs(nestedList);
        iterator = list.iterator();
    }

    @Override
    public Integer next() {
        return iterator.next();
    }

    @Override
    public boolean hasNext() {
        return iterator.hasNext();
    }

    private void dfs(List<NestedInteger> l) {
        for (NestedInteger nestedInteger : l) {
            if (nestedInteger.isInteger()) {
                list.add(nestedInteger.getInteger());
            } else {
                dfs(nestedInteger.getList());
            }
        }
    }
}

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