Zigzag Iterator II 541

Question

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

Example

Given k = 3 1d vectors:

[1,2,3]

[4,5,6,7]

[8,9]

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

Solution

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）。

代码如下：

public class ZigzagIterator2 {
    /**
     * @param vecs a list of 1d vectors
     */
    //用iterator更省空间
    private ArrayList<Iterator<Integer>> vec;
    private int count;
    public ZigzagIterator2(ArrayList<ArrayList<Integer>> vecs) {
        // initialize your data structure here.
        vec = new ArrayList<Iterator<Integer>>();
        for(ArrayList<Integer> list : vecs){
            if(list.size() > 0){
                vec.add(list.iterator());
            }
        }
        count = 0;
    }

    public int next() {
        // Write your code here
        int res = vec.get(count).next();
        if(vec.get(count).hasNext()){
            count = (count + 1) % vec.size();
        }else{
            vec.remove(count);
            if(vec.size() > 0){
                count %= vec.size();
            }
        }
        return res;
    }

    public boolean hasNext() {
        // Write your code here
        return vec.size() > 0;
    }
}

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