Zigzag Iterator

最新推荐文章于 2021-08-12 21:54:08 发布

原创最新推荐文章于 2021-08-12 21:54:08 发布 · 360 阅读

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本文介绍了一种交错迭代器的实现方法，用于返回多个一维向量元素的交错序列。通过具体示例展示了如何使用该迭代器，并探讨了其在面对更多输入向量时的扩展性。

Problem

Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

v1 = [1, 2]
v2 = [3, 4, 5, 6]

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

Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

[1,2,3]
[4,5,6,7]
[8,9]

It should return [1,4,8,2,5,9,3,6,7].

Solution

关键是hasNext()如何实现。

大概思路是用curRow 和 curCol来记录当前位置，curRow从0到K－1，当curRow到 K时， curCol就加一。

细节是何时停止？每次curRow 从0 开始时，假设该列为空（assumeEmpty = true），当curRow不断加一时，如果遇到一个有效数字，就将assumeEmpty改为false。

当curRow一直到底而且assumeEmpty还是为空时，那么就说明到底了；否则，curRow清零，curCol 加一，重新调用 hasNext();

class ZigzagIterator {
    int K;
    int curRow;
    int curCol;
    bool isEmpty;
    vector<vector<int>> matrix;
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        matrix.push_back(v1);
        matrix.push_back(v2);
        
        K = 2;
        curRow = 0;
        curCol = 0;
        isEmpty = true;
    }

    int next() {
        return matrix[curRow++][curCol];
    }

    bool hasNext() {
        if(curRow == 0) isEmpty = true;
        
        while(curRow < K) {
            if(matrix[curRow].size() > curCol) {
                isEmpty = false;
                return true;
            }
            curRow++;
        }
        
        if(isEmpty) return false;
        
        curRow = 0;
        curCol++;
        return hasNext();
    }
};

/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i(v1, v2);
 * while (i.hasNext()) cout << i.next();
 */