題目

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

這標題，翻譯的是人話嗎？啥叫扁平化嵌套列表迭代器？。。

題解

比較像深度優先搜索。思路是：先 flatten 成 list 存好，需要的時候直接返回。感覺沒有 get 到這個問題的精髓。評論區也有人質疑這種方法：

關于這個問題，我一開始也沒打算先存成 list，僅存儲調用的指針，沒寫出來，就先用了個 list 存數據了。

/*** // 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());}}}@Overridepublic Integer next() {if (index < list.size()) return list.get(index++);else return null;}@Overridepublic 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();*/

總結

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