[leetcode] 251. Flatten 2D Vector 解题报告_leetcode 251 flatten 2d vector-优快云博客

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/50913164

这篇博客详细解析了LeetCode中的251题，要求实现一个迭代器来扁平化二维向量。文章讨论了如何处理空行，并强调了保持两个变量的有效性作为不变量的重要性，同时提出了重构重复逻辑到公共方法的建议。作者认为虽然题目要求仅用两个变量，但在Java中可能需要三个迭代器来完成任务。

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题目链接: https://leetcode.com/problems/flatten-2d-vector/

Implement an iterator to flatten a 2d vector.

For example,
Given 2d vector =

[
  [1,2],
  [3],
  [4,5,6]
]

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,2,3,4,5,6].

Hint:

How many variables do you need to keep track?
Two variables is all you need. Try with x and y.
Beware of empty rows. It could be the first few rows.
To write correct code, think about the invariant to maintain. What is it?
The invariant is x and y must always point to a valid point in the 2d vector. Should you maintain your invariant ahead of time or right when you need it?
Not sure? Think about how you would implement hasNext(). Which is more complex?
Common logic in two different places should be refactored into a common method.

Follow up:
As an added challenge, try to code it using only iterators in C++ or iterators in Java.

思路: 可以维护三个迭代器, 感觉要求说只要维护两个遍历有点不太可能. java或许可以.

代码各自如下:

class Vector2D {
public:
    Vector2D(vector<vector<int>>& vec2d) {
        if(vec2d.size() == 0) return;
        it1 = vec2d.begin();
        itEnd = vec2d.end();
        it2 = (*it1).begin();
    }

    int next() {
        return *it2++;
    }

    bool hasNext() {
        while(it1 != itEnd && it2 == (*it1).end())
        {
            it1++;
            it2 = (*it1).begin();
        }
        return it1 != itEnd;
    }
private:
    vector<vector<int>>::iterator it1, itEnd;
    vector<int>::iterator it2;
};

/**
 * Your Vector2D object will be instantiated and called as such:
 * Vector2D i(vec2d);
 * while (i.hasNext()) cout << i.next();
 */