[LeetCode] Flatten Nested List Iterator 压平嵌套链表迭代器

 

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].

 

这道题让咱们创建压平嵌套链表的迭代器，关于嵌套链表的数据结构最先出如今Nested List Weight Sum中，而那道题是用的递归的方法来解的，而迭代器通常都是用迭代的方法来解的，而递归通常都需用栈来辅助遍历，因为栈的后进先出的特性，咱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，咱们的hasNext()函数须要遍历栈，并进行处理，若是栈顶元素是整数，直接返回true，若是不是，那么移除栈顶元素，并开始遍历这个取出的list，仍是从后往前压入栈，循环中止条件是栈为空，返回false，参见代码以下：

 

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;
};

 

咱们也可使用deque来代替stack，实现思路和上面彻底同样，参见代码以下:

 

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;
};

 

虽然说迭代器是要用迭代的方法，可是咱们能够强行使用递归来解，怎么个强行法呢，就是咱们使用一个队列queue，在构造函数的时候就利用迭代的方法把这个嵌套链表所有压平展开，而后在调用hasNext()和next()就很简单了：

 

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());
        }
    }
};

 

相似题目：

Nested List Weight Sum

Flatten 2D Vector

Zigzag Iterator

 

参考资料：

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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