题目

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].

用两个Iterator<Integer>

代码

public class ZigzagIterator {

private static Iterator<Integer> ite1 = null, ite2 = null;
private static boolean turnIte1 = false;

public ZigzagIterator(List<Integer> list1, List<Integer> list2) {
ite1 = list1.iterator();
ite2 = list2.iterator();
turnIte1 = true;
}
public boolean hasNext() {
return ite1.hasNext() || ite2.hasNext();
}
public int next() {
Iterator<Integer> localIte = (turnIte1)? ite1 : ite2;
if (hasNext()) {
if ((localIte == ite1) && (!ite1.hasNext())) { localIte = ite2; }
if ((localIte == ite2) && (!ite2.hasNext())) { localIte = ite1; }
}
turnIte1 = !turnIte1;
return localIte.next();
}
}

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

为了适应多个数列，可以用一个List<Iterator<Integer>>

代码

public class ZigzagIterator {

private List<Iterator<Integer>> ites = new ArrayList<>();

public ZigzagIterator(List<Integer> list1, List<Integer> list2) {
Iterator<Integer> ite = list1.iterator();
ite = list2.iterator();
}
public boolean hasNext() {
return !ites.isEmpty();
}
public int next() {
if (hasNext()) {
Iterator<Integer> localIte = ites.remove(0);
Integer res = localIte.next();
if (localIte.hasNext()) {
}
return res;
}
return 0;
}

}

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