1、题目描述
https://leetcode-cn.com/problems/flatten-nested-list-iterator/
2、解法
利用多叉树遍历的方法,把所有的值取出按顺序放到一个result列表里面,然后迭代器就取该result list的迭代器实现。
/**
* // 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 Iterator<Integer> iterator;
public NestedIterator(List<NestedInteger> nestedList) {
//遍历所有的节点
List<Integer> result = new LinkedList<>();
for(NestedInteger node:nestedList){
traverse(node,result);
}
iterator = result.iterator();//获取每个nestedList的迭代器
}
@Override
public Integer next() {
return iterator.next();
}
@Override
public boolean hasNext() {
return iterator.hasNext();//利用其对应的迭代器
}
private void traverse(NestedInteger root,List<Integer>result){
//N叉树的遍历
if(root.isInteger())//是整数,则是叶子节点
{
result.add(root.getInteger());
}
//遍历
for(NestedInteger child:root.getList()){
traverse(child,result);
}
}
}
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i = new NestedIterator(nestedList);
* while (i.hasNext()) v[f()] = i.next();
*/
运行结果:
但是上面是我们访问一个元素,就要遍历整颗树的所有结点,把叶子结点的值添加到result。如果规模变大,那么构造器构造时间很长,效率低下。而我们知道迭代器应该是懒加载的。每次取一部分。
进阶解法
我们对每个这样的数据结构进行判断
- 如果其不是整数就说明它是list,则将其子元素(可能为整数也可能为list)重新加入一个总列表里面,其自身则移除出去
- 如果是整数就直接返回
!list.isEmpty();表明下一次是否可以取数
使得该列表总可以保持
/**
* // 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 LinkedList<NestedInteger> nodeList;
public NestedIterator(List<NestedInteger> nestedList) {
nodeList = new LinkedList(nestedList);//这里使用链式列表
}
@Override
public Integer next() {
//hasNext()保证了加入该list的都是整数
return nodeList.remove(0).getInteger();
}
@Override
public boolean hasNext() {
while(!nodeList.isEmpty()&& !nodeList.get(0).isInteger()){
//是列表而不是整数
//是列表,其不含整数,则把它的子列表取出来,加入链表头中
List<NestedInteger> first = nodeList.remove(0).getList();
for(int i=first.size()-1;i>=0;i--){
//遍历其所有的值
nodeList.addFirst(first.get(i));
}
//if(first==null) continue;//取下一个
}
return !nodeList.isEmpty();//空则说明上面是因为该数不是整数,则说明我们不能取了
}
}
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i = new NestedIterator(nestedList);
* while (i.hasNext()) v[f()] = i.next();
*/
关于这个效率为什么反而变低了呢?因为数量比较少的情况,直接遍历全部比较快,而使用第二种方式,每次都要执行hashNext里面的大段代码。估计是懒加载累积导致性能变差。