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]
.
一刷
题解: 用一个queue装两个list的iterator
public class ZigzagIterator {
Queue<Iterator<Integer>> list;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
list = new LinkedList<>();
if(!v1.isEmpty()) list.add(v1.iterator());
if(!v2.isEmpty()) list.add(v2.iterator());
}
public int next() {
Iterator poll = list.poll();
int res = (Integer)poll.next();
if(poll.hasNext()) list.add(poll);
return res;
}
public boolean hasNext() {
return !list.isEmpty();
}
}
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2);
* while (i.hasNext()) v[f()] = i.next();
*/
二刷
怎么比queue快一倍。用flag表示是从第一个还是第二个取
public class ZigzagIterator {
int v1Index;
int v2Index;
List<Integer> v1;
List<Integer> v2;
boolean flag;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
this.v1 = v1;
this.v2 = v2;
v1Index = 0;
v2Index = 0;
flag = true;
}
public int next() {
if (flag) {
flag = false;
return v1.get(v1Index++);
} else {
flag = true;
return v2.get(v2Index++);
}
}
public boolean hasNext() {
if (v1Index >= v1.size() && v2Index >= v2.size()) {
return false;
}
if (flag) {
if (v1Index < v1.size()) {
return true;
} else {
flag = false;
return hasNext();
}
} else {
if (v2Index < v2.size()) {
return true;
} else {
flag = true;
return hasNext();
}
}
}
}
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2);
* while (i.hasNext()) v[f()] = i.next();
*/