[抄题]:
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?
Clarification for 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]
.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
不用写ifv1 if v2.hasnext return true; 直接用temp交换就行了
[英文数据结构或算法,为什么不用别的数据结构或算法]:
不知道初始化要写变量 private Iterator<Integer> i, j, tmp;
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 交换的条件是:i.hasNext(), 别人有下一个才能和你换
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
实现类的题目不能设为public class,否则要改文件名。
[潜台词] :
// package whatever; // don't place package name! import java.io.*; import java.util.*; import java.lang.*; /* Input: v1 = [1,2] v2 = [3,4,5,6] Output: [1,3,2,4,5,6] */ class driverFuction { public static void main (String[] args) { List<Integer> v1 = Arrays.asList(1, 2); List<Integer> v2 = Arrays.asList(3, 4, 5, 6); ZigzagIterator answer = new ZigzagIterator(v1, v2); System.out.println(answer.next()); System.out.println(answer.next()); System.out.println(answer.next()); System.out.println(answer.next()); System.out.println(answer.next()); } } class ZigzagIterator { Iterator<Integer> i, j, temp; public ZigzagIterator(List<Integer> v1, List<Integer> v2) { i = v1.iterator(); j = v2.iterator(); } public int next() { if (i.hasNext()) {temp = j; j = i; i = temp;} return j.next(); } public boolean hasNext() { if (i.hasNext() || j.hasNext()) return true; return false; } }