leetcode 341. Flatten Nested List Iterator

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]

.

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很显然这个是一个递归问题，但是题目要求实现一个迭代器，这里我想到两个方法：
1. 直接在构造函数中使用递归构造好整个序列，然后逐个返回
2.栈保存历史记录，栈顶保存当前打印的列表以及打印到列表哪个位置，刚开始实现很顺利，但是后面提交发现测试用例有空列表的情况，如输入为[[],[[]]]，这个
就麻烦了，该进后的方式就是保存一个缓存节点
下面是自己实现的方法：

#include <stack>
#include <vector>
#include <assert.h>
struct NestedContext {
const vector<NestedInteger>* vec;
size_t pos;
};

class NestedIterator {
public:
NestedIterator(vector<NestedInteger> &nestedList)
{
has_next = true;
if (nestedList.empty()) {
return;
}
NestedContext ctx = {&nestedList, 0};
s.push(ctx);
}
int next() {
return cache_val;
}

bool _next(int& val) {
while (!s.empty()) {
NestedContext& ctx = s.top();
if (ctx.vec->at(ctx.pos).isInteger()) {
int rv = ctx.vec->at(ctx.pos++).getInteger();
if (ctx.pos >= ctx.vec->size()) {
s.pop();
}
val = rv;
return true;
}
else {
const vector<NestedInteger>& nestedList = ctx.vec->at(ctx.pos++).getList();
if (ctx.pos >= ctx.vec->size()) {
s.pop();
}
if (!nestedList.empty()) {
NestedContext new_ctx = {&nestedList, 0};
s.push(new_ctx);
}
}
}
return false;
}

bool hasNext() {
has_next = _next(cache_val);
return !s.empty() || has_next;
}
stack<NestedContext> s;
bool has_next;
int cache_val;
};