281. Zigzag Iterator

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

Example:

Input:

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


Output:
[1,3,2,4,5,6]


Explanation:
 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?

Clarificationfor the follow up question:
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:

Input:

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


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

Solution

(1) Java

public class ZigzagIterator {

    Deque<Iterator<Integer>> iterq = new LinkedList<>();

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        if (v1.size() > 0) {
            iterq.offer(v1.iterator());
        }
        if (v2.size() > 0) {
            iterq.offer(v2.iterator());
        }
    }

    public int next() {
        Iterator<Integer> iter = iterq.poll();
        int rst = iter.next();
        if (iter.hasNext()) {
            iterq.offer(iter);
        }
        return rst;
    }

    public boolean hasNext() {
        return !iterq.isEmpty();
    }
}

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

(2) Python



(3) Scala



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