Zigzag Iterator

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].

Tips:

• 对iterator的定义和写法不是很清晰，算法倒是比较简洁明了。借鉴了discuss做出来，但是并不是很难。

• 两个数组的合并，穿插进行。

• Follow up 要新建arraylist iterator。

//解法1：

public class ZigzagIterator {
    Iterator<Integer> it1;
    Iterator<Integer> it2;
    int turns;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        this.it1 = v1.iterator();
        this.it2 = v2.iterator();
        turns = 0;
    }

    public int next() {
        // 如果没有下一个则返回0
        if(!hasNext()){
            return 0;
        }
        turns++;
        // 如果是第奇数个，且第一个列表也有下一个元素时，返回第一个列表的下一个
        // 如果第二个列表已经没有，返回第一个列表的下一个
        if((turns % 2 == 1 && it1.hasNext()) || (!it2.hasNext())){
            return it1.next();
        // 如果是第偶数个，且第二个列表也有下一个元素时，返回第二个列表的下一个
        // 如果第一个列表已经没有，返回第二个列表的下一个
        } else if((turns % 2 == 0 && it2.hasNext()) || (!it1.hasNext())){
            return it2.next();
        }
        return 0;
    }

    public boolean hasNext() {
        return it1.hasNext() || it2.hasNext();
    }
}

//解法2：

public class ZigzagIterator {
int index;
    boolean count;
    List<Integer> v1;
    List<Integer> v2;

    public ZigzagIterator (List<Integer> v1, List<Integer> v2) {
        this.index = 0;
        this.count = true;
        this.v1 = v1;
        this.v2 = v2;
    }

    public int next() {
        int result;
        if (count && index < v1.size()) {
            result = v1.get(index);
            if (index < v2.size()) {
                count = !count;
            } else {
                index++;
            }
        } else {
            result = v2.get(index++);
            if (index < v1.size()) {
                count = !count;
            }
        }
        return result;
    }

    public boolean hasNext() {
        return index < v1.size() || index < v2.size();
    }
}

/**

• Your ZigzagIterator object will be instantiated and called as such:
• ZigzagIterator i = new ZigzagIterator(v1, v2);
• while (i.hasNext()) v[f()] = i.next(); */

## follow up:

public class ZigzagIterator implements Iterator<Integer> {    
    List<Iterator<Integer>> itlist;
    int turns;

    public ZigzagIterator(List<Iterator<Integer>> list) {
        this.itlist = new LinkedList<Iterator<Integer>>();
        // 将非空迭代器加入列表
        for(Iterator<Integer> it : list){
            if(it.hasNext()){
                itlist.add(it);
            }
        }
        turns = 0;
    }

    public Integer next() {
        if(!hasNext()){
            return 0;
        }
        Integer res = 0;
        // 算出本次使用的迭代器的下标
        int pos = turns % itlist.size();
        Iterator<Integer> curr = itlist.get(pos);
        res = curr.next();
        // 如果这个迭代器用完，就将其从列表中移出
        if(!curr.hasNext()){
            itlist.remove(turns % itlist.size());
            // turns变量更新为上一个下标
            turns = pos - 1;
        }
        turns++;
        return res;
    }

    public boolean hasNext() {
        return itlist.size() > 0;
    }
}