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".Example1
Input: k = 3
vecs = [
[1,2,3],
[4,5,6,7],
[8,9],
]
Output: [1,4,8,2,5,9,3,6,7]
Example2
Input: k = 3 vecs = [ [1,1,1] [2,2,2] [3,3,3] ] Output: [1,2,3,1,2,3,1,2,3]
思路：
用k个指针维护每个数组的当前最靠前的元素，循环枚举这k个指针，每次枚举到就在答案序列中将该数加入，如果指针到了该数组的尾部，那么就不作处理
public class ZigzagIterator2 {
public List<Iterator<Integer>> its;
public int turns;
/**
* @param vecs a list of 1d vectors
*/
public ZigzagIterator2(List<List<Integer>> vecs) {
// initialize your data structure here.
this.its = new ArrayList<Iterator<Integer>>();
for (List<Integer> vec : vecs) {
if (vec.size() > 0)
its.add(vec.iterator());
}
turns = 0;
}
public int next() {
// Write your code here
int elem = its.get(turns).next();
if (its.get(turns).hasNext())
turns = (turns + 1) % its.size();
else {
its.remove(turns);
if (its.size() > 0)
turns %= its.size();
}
return elem;
}
public boolean hasNext() {
// Write your code here
return its.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
*/
原文：https://www.cnblogs.com/FLAGyuri/p/12078563.html