LintCode 540. Zigzag Iterator 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/zigzag-iterator/
Given two 1d vectors, implement an iterator to return their elements alternately.
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]
解题思路
题目是设计一个穿梭遍历两个数组的iterator。
- 关键点：三个变量。分别遍历list1，list2的iterator变量以及决定下一个数是从v1还是从v2里拿的turns变量。
参考代码
public class ZigzagIterator {
/**
* @param v1 v2 two 1d vectors
*/
public Iterator<Integer> it1;
public Iterator<Integer> it2;
public int turns;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
// initialize your data structure here.
this.it1 = v1.iterator();
this.it2 = v2.iterator();
turns = 0;
}
public int next() {
turns++;
if ((turns % 2 == 1 && it1.hasNext()) || (!it2.hasNext())) {
return it1.next();
} else if ((turns % 2 == 0 && it2.hasNext()) || (!it1.hasNext())) {
return it2.next();
} else {
return -1;
}
}
public boolean hasNext() {
return it1.hasNext() || it2.hasNext();
}
}
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator solution = new ZigzagIterator(v1, v2);
* while (solution.hasNext()) result.add(solution.next());
* Output result
*/