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題目截圖

public class ZigzagIterator {
    private Iterator<Integer> iter1, iter2;
    private Iterator<Integer> iter;
    
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        iter1 = v1.iterator();
        iter2 = v2.iterator();
        
        // pay attention here. list1 could be empty
        iter = iter1.hasNext() ? iter1 : iter2; 
    }

    public int next() {
        int nextInt = iter.next();
        
        if (iter == iter1 && iter2.hasNext()) {
            iter = iter2;
        } else if (iter == iter2 && iter1.hasNext()) {
            iter = iter1;
        }
        
        return nextInt;
    }

    public boolean hasNext() {
        return iter.hasNext();
    }
}

Follow up: Extend to List of List

List of iterators

1. 如果保持所有的 iterators，那么 next() 和 hasNext() can be up to O(n) runtime, n is the size of list of lists. 因為隨著 next() 的使用，List of iterators 中會有 iter does not have next.
2. 如果只是保持有 next 的 iterators，那么 next() 和 hasNext() 將都是 O(1) runtime. 需要 LinkedList<Iterator<Integer>> 的首個 iter 即為 nextIterator。

public class ZigzagIterator {
    private LinkedList<Iterator<Integer>> iterators;
    
    public ZigzagIterator(List<List<Integer>> lists) {
        iterators = new LinkedList<>();
        
        for (List<Integer> list: lists) {
            if (list != null && !list.isEmpty()) {
                iterators.add(list.iterator());
            }
        }
    }
    
    public int next() {
        Iterator<Integer> iter = iterators.removeFirst();
        int nextInt = iter.next();
        
        if (iter.hasNext()) {
            iterators.add(iter);
        }
        
        return nextInt;
    }
    
    public boolean hasNext() {
        return !iterators.isEmpty();
    } 
}