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].

 

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:

 

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;
};

 

我们也可以使用deque来代替stack,实现思路和上面完全一样,参见代码如下:

 

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:

 

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());
        }
    }
};

 

类似题目:

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

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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