[LeetCode] 341. Flatten Nested List Iterator Java

题目：

Given a nested list of integers, implement an iterator to flatten it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Example 1:
Given the list [[1,1],2,[1,1]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].

Example 2:
Given the list [1,[4,[6]]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].

题意及分析：嵌套的list，要求给出其迭代器的next()和hasNext函数。可以使用一个stack来维护当前list剩余元素。我开始是在next里对stack做操作，但是由于当出现[]这种空list时没有办法判断hasNext，容易出现空指针操作。所以需要在hasNext()里面对stack操作。将list反向添加进stack中，这样就可以让先出现的数先被stack弹出。

代码：

public class NestedIterator implements Iterator<Integer> {

    List<NestedInteger> list;
    Stack<NestedInteger> stack = new Stack<>();
    NestedInteger nowInteger;
    public NestedIterator(List<NestedInteger> nestedList) {
        list = nestedList;
        for(int i=list.size()-1;i>=0;i--){//对每一个list都反转
            stack.push(list.get(i));
        }
    }

    @Override
    public Integer next() {
        return stack.pop().getInteger();
    }

    @Override
    public boolean hasNext() {
        if(stack.isEmpty())
            return false;
        else{
            nowInteger = stack.peek();
            if(nowInteger.isInteger()){
                return true;
            }else{
                while(!stack.isEmpty()){
                    nowInteger=stack.pop();
                    List<NestedInteger> list = nowInteger.getList();
                    for(int i=list.size()-1;i>=0;i--){
                        stack.push(list.get(i));
                    }
                    if(stack.isEmpty()) return false;   //如果没有next直接返回false
                    nowInteger = stack.peek();      //取出栈顶元素
                    if(nowInteger.isInteger())
                        return true;
                }
            }
        }
        return false;
    }
}