[LeetCode] Flatten Nested List Iterator 壓平嵌套鏈表迭代器

 

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].

 

這道題讓咱們創建壓平嵌套鏈表的迭代器，關於嵌套鏈表的數據結構最先出如今Nested List Weight Sum中，而那道題是用的遞歸的方法來解的，而迭代器通常都是用迭代的方法來解的，而遞歸通常都需用棧來輔助遍歷，因爲棧的後進先出的特性，咱們在對向量遍歷的時候，從後往前把對象壓入棧中，那麼第一個對象最後壓入棧就會第一個取出來處理，咱們的hasNext()函數須要遍歷棧，並進行處理，若是棧頂元素是整數，直接返回true，若是不是，那麼移除棧頂元素，並開始遍歷這個取出的list，仍是從後往前壓入棧，循環中止條件是棧爲空，返回false，參見代碼以下：

 

解法一:

class NestedIterator {
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();
            for (int i = t.getList().size() - 1; i >= 0; --i) {
                s.push(t.getList()[i]);
            }
        }
        return false;
    }

private:
    stack<NestedInteger> s;
};

 

咱們也可使用deque來代替stack，實現思路和上面徹底同樣，參見代碼以下:

 

解法二：

class NestedIterator {
public:
    NestedIterator(vector<NestedInteger> &nestedList) {
        for (auto a : nestedList) {
            d.push_back(a);
        }
    }

    int next() {
        NestedInteger t = d.front(); d.pop_front();
        return t.getInteger();
    }

    bool hasNext() {
        while (!d.empty()) {
            NestedInteger t = d.front();
            if (t.isInteger()) return true;
            d.pop_front();
            for (int i = 0; i < t.getList().size(); ++i) {
                d.insert(d.begin() + i, t.getList()[i]);
            }
        }
        return false;
    }

private:
    deque<NestedInteger> d;
};

 

雖然說迭代器是要用迭代的方法，可是咱們能夠強行使用遞歸來解，怎麼個強行法呢，就是咱們使用一個隊列queue，在構造函數的時候就利用迭代的方法把這個嵌套鏈表所有壓平展開，而後在調用hasNext()和next()就很簡單了：

 

解法三：

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) {
        for (auto a : nestedList) {
            if (a.isInteger()) q.push(a.getInteger());
            else make_queue(a.getList());
        }
    }
};

 

相似題目：

Nested List Weight Sum

Flatten 2D Vector

Zigzag Iterator

 

參考資料：

https://leetcode.com/discuss/95841/simple-solution-with-queue

https://leetcode.com/discuss/95892/concise-c-without-storing-all-values-at-initialization

 

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