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();
*/