给定一个列表，该列表中的每个要素要么是个列表，要么是整数。将其变成一个只包含整数的简单列表。

易错点：nestedList.get(i).getInteger()，取集合中的元素时忘记get(i),,就取不到了

扩展：怎么用非递归来解答

思路：比较简单，直接递归调用即可。

/**
 * // 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 Solution {

    // @param nestedList a list of NestedInteger
    // @return a list of integer
    public List<Integer> flatten(List<NestedInteger> nestedList) {
        // Write your code here
        List<Integer> list=new ArrayList<>();
        doFlatten(nestedList,list);
        return list;
    }
     public void doFlatten(List<NestedInteger> nestedList,List<Integer> list){
         if(nestedList != null){
             for(int i=0;i<nestedList.size();i++){
                 if(nestedList.get(i).isInteger()){
                     list.add(nestedList.get(i).getInteger());
                 }else{
                     doFlatten(nestedList.get(i).getList(),list);
                 }
             }
         }
     }
}