541.Zigzag Iterator II

1.Description(Medium)

Follow up Zigzag Iterator: What if you are givenk1d vectors? How well can your code be extended to such cases? The "Zigzag" order is not clearly defined and is ambiguous fork > 2cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic".

Example

Givenk = 31d vectors:

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

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

2.Code

public class ZigzagIterator2 {
    /**
     * @param vecs a list of 1d vectors
     */
    public ZigzagIterator2(ArrayList<ArrayList<Integer>> vecs) {
        // initialize your data structure here.
    }

    public int next() {
        // Write your code here
    }

    public boolean hasNext() {
        // Write your code here   
    }
}

/**
 * Your ZigzagIterator2 object will be instantiated and called as such:
 * ZigzagIterator2 solution = new ZigzagIterator2(vecs);
 * while (solution.hasNext()) result.add(solution.next());
 * Output result
 */

和I类似，只是用一个list来存k个iterator（如果iterator中没有元素则不用加入）。用一个count%list_size来确定应该用哪个iterator。若其中一个iterator里面的数已经被取完（即该iterator的hasNext()为false），则将其从list中移除，并将count设置为count%new list_size（如果new list_size != 0）。

public class Solution_541 {
    /**
     * @param vecs a list of 1d vectors
     */

    private List<Iterator<Integer>> iteratorList;
    private int index;

    public Solution_541(ArrayList<ArrayList<Integer>> vecs) {
        this.iteratorList=new ArrayList<Iterator<Integer>>();
        for(ArrayList<Integer> list:vecs){
            if(list.size()>0){
                iteratorList.add(list.iterator());
            }
        }
        index=0;
    }

    public int next() {
        int current=iteratorList.get(index).next();
        if(iteratorList.get(index).hasNext()){
            index=(index+1)%iteratorList.size();
        }else//若其中一个iterator里面的数已经被取完（即该iterator的hasNext()为false），
            //则将其从list中移除，并将count设置为count%new list_size（如果new list_size != 0）。
        {
            iteratorList.remove(index);
            if(iteratorList.size()>0){
                int newsize=iteratorList.size();
                index=index%newsize;
            }
        }
        return current;
    }

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