Code & Func

第100天。

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].

挺有趣的题目，主要是要实现一个嵌套列表的迭代器，大概是三个函数：

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class NestedIterator {
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public:
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    NestedIterator(vector<NestedInteger> &nestedList) {
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    }
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    int next() {
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    }
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    bool hasNext() {
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    }
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};

然后他也提供了NestedInteger的接口和一些说明，算是对题目的补充：

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/**
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 * // This is the interface that allows for creating nested lists.
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 * // You should not implement it, or speculate about its implementation
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 * class NestedInteger {
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 *   public:
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 *     // Return true if this NestedInteger holds a single integer, rather than a nested list.
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 *     bool isInteger() const;
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 *
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 *     // Return the single integer that this NestedInteger holds, if it holds a single integer
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 *     // The result is undefined if this NestedInteger holds a nested list
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 *     int getInteger() const;
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 *
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 *     // Return the nested list that this NestedInteger holds, if it holds a nested list
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 *     // The result is undefined if this NestedInteger holds a single integer
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 *     const vector<NestedInteger> &getList() const;
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 * };
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 */
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/**
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 * Your NestedIterator object will be instantiated and called as such:
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 * NestedIterator i(nestedList);
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 * while (i.hasNext()) cout << i.next();
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 */

从补充中我们可以看到，在调用next前一定会先调用hasNext,有了这个前提我们写起来会方便一点。

我的解法是，NestedIterator只保存构造函数中传入的nestedList的两个迭代器，之所以是两个，是因为要保存end迭代器，然后要实现嵌套，我们还要一个NestedIterator的指针，利用这个指针来对下一级列表的元素进行迭代：

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class NestedIterator {
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public:
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    NestedIterator(vector<NestedInteger> &nestedList) {
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        _it = nestedList.begin(); _end_it = nestedList.end();
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        _tmp_it = nullptr;
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    }
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    int next() {
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        if (_it->isInteger()) { int ret = _it->getInteger(); ++_it; return ret; }
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        return _tmp_it->next();
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    }
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    bool hasNext() {
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        if (_it == _end_it) return false;
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        if (_it->isInteger()) return true;
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        if (_tmp_it == nullptr) {
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            _tmp_it = new NestedIterator(_it->getList());
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        }
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        if (_tmp_it->hasNext()) return true;
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        delete _tmp_it; _tmp_it = nullptr; ++_it;
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        return hasNext();
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    }
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private:
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    vector<NestedInteger>::iterator _it, _end_it;
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    NestedIterator *_tmp_it;
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};

然后这里的实现虽然比较简单，简洁，但是在遇到一些特殊情况的时候会对性能造成极大的影响,比如说[[[[[1,2,3]]]]],虽然只有三个元素，但是因为有5层的嵌套，我们要有5个迭代器，每次调用next和hasNext都需要递归调用5次才能返回，这样效率就有点低了.

dicuss中的解法会比较好一点，类别DFS来做，先用stack保存所有的元素，在调用hasNext的时候，如果栈顶是列表就将其展开并压栈（倒序），然后在递归调用hasNext，直到栈顶为数字时,然后调用next就直接返回栈顶即可：

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class NestedIterator {
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public:
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    NestedIterator(vector<NestedInteger> &nestedList) {
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        begins.push(nestedList.begin());
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        ends.push(nestedList.end());
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    }
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    int next() {
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        hasNext();
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        return (begins.top()++)->getInteger();
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    }
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    bool hasNext() {
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        while (begins.size()) {
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            if (begins.top() == ends.top()) {
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                begins.pop();
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                ends.pop();
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            } else {
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                auto x = begins.top();
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                if (x->isInteger())
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                    return true;
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                begins.top()++;
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                begins.push(x->getList().begin());
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                ends.push(x->getList().end());
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            }
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        }
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        return false;
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    }
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private:
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    stack<vector<NestedInteger>::iterator> begins, ends;
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};

update at 2020-04-03

和第二个解法有点像，但是只需要一个栈即可：

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class NestedIterator {
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public:
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    stack<vector<NestedInteger>::iterator> st;
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    NestedIterator(vector<NestedInteger> &nestedList) {
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        reversePush(nestedList);
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    }
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    void reversePush(vector<NestedInteger> &nestedList) {
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        auto beg = nestedList.begin();
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        for(int i = nestedList.size() - 1; i >= 0; --i) {
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            st.push(beg + i);
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        }
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    }
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    int next() {
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        auto top = st.top(); st.pop();
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        return top->getInteger();
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    }
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    bool hasNext() {
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        while(!st.empty() && !st.top()->isInteger()) {
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            auto top = st.top(); st.pop();
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            reversePush(top->getList());
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        }
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        return !st.empty();
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    }
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};