341. Flatten Nested List Iterator (Medium)

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].

Solution 1: Stack

这道题让我们建立压平嵌套链表的迭代器，关于嵌套链表的数据结构最早出现在Nested List Weight Sum中，而那道题是用的递归的方法来解的，而迭代器一般都是用迭代的方法来解的，而递归一般都需用栈来辅助遍历，由于栈的后进先出的特性，我们在对向量遍历的时候，从后往前把对象压入栈中，那么第一个对象最后压入栈就会第一个取出来处理，我们的hasNext()函数需要遍历栈，并进行处理，如果栈顶元素是整数，直接返回true，如果不是，那么移除栈顶元素，并开始遍历这个取出的list，还是从后往前压入栈，循环停止条件是栈为空，返回false，参见代码如下：

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * class NestedInteger {
 *   public:
 *     // Return true if this NestedInteger holds a single integer, rather than a nested list.
 *     bool isInteger() const;
 *
 *     // Return the single integer that this NestedInteger holds, if it holds a single integer
 *     // The result is undefined if this NestedInteger holds a nested list
 *     int getInteger() const;
 *
 *     // Return the nested list that this NestedInteger holds, if it holds a nested list
 *     // The result is undefined if this NestedInteger holds a single integer
 *     const vector<NestedInteger> &getList() const;
 * };
 */
class NestedIterator {
    int i;
public:
    NestedIterator(vector<NestedInteger> &nestedList) {
        for (int i = nestedList.size()-1; i >= 0; --i) {
            s.push(nestedList[i]);
        }
    }

    int next() {
        NestedInteger t = s.top(); s.pop();
        return t.getInteger();
    }

    bool hasNext() {
        while (!s.empty()) {
            NestedInteger t = s.top();
            if (t.isInteger()) return true;
            s.pop();
            const vector<NestedInteger> &tmp = t.getList();
            for (int i = tmp.size()-1; i >= 0; --i) {
                s.push(tmp[i]);
            }
        }
        return false;
    }
private:
    stack<NestedInteger> s;
};

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i(nestedList);
 * while (i.hasNext()) cout << i.next();
 */

Solution 2: Queue, recursive

虽说迭代器是要用迭代的方法，但是我们可以强行使用递归来解，怎么个强行法呢，就是我们使用一个队列queue，在构造函数的时候就利用迭代的方法把这个嵌套链表全部压平展开，然后在调用hasNext()和next()就很简单了：

class NestedIterator {
public:
    NestedIterator(vector<NestedInteger> &nestedList) {
        make_queue(nestedList);
    }

    int next() {
        int t = q.front(); q.pop();
        return t;
    }

    bool hasNext() {
        return !q.empty();
    }
private:
    queue<int> q;
    void make_queue(vector<NestedInteger> &nestedList) {
        if (nestedList.empty()) return;
        for (auto &t: nestedList) {
            if (t.isInteger()) q.push(t.getInteger());
            else make_queue(t.getList());
        }
    }
};