[LeetCode] 281. Zigzag Iterator

 

Given two vectors of integers v1 and v2, implement an iterator to return their elements alternately.

Implement the ZigzagIterator class:

• ZigzagIterator(List<int> v1, List<int> v2) initializes the object with the two vectors v1 and v2.
• boolean hasNext() returns true if the iterator still has elements, and false otherwise.
• int next() returns the current element of the iterator and moves the iterator to the next element.

Example 1:

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].

Example 2:

Input: v1 = [1], v2 = []
Output: [1]

Example 3:

Input: v1 = [], v2 = [1]
Output: [1]

Constraints:

• 0 <= v1.length, v2.length <= 1000
• 1 <= v1.length + v2.length <= 2000
• -231 <= v1[i], v2[i] <= 231 - 1

Follow up: What if you are given k 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".

Follow-up Example:

Input: v1 = [1,2,3], v2 = [4,5,6,7], v3 = [8,9]
Output: [1,4,8,2,5,9,3,6,7]

锯齿迭代器。

题意是给两个一维的向量，请交替输出他们的值。

常规思路就是为这两个向量分别创建一个iterator迭代器，然后从两个迭代器里交替输出元素。这里需要一个temp迭代器以用来帮助交换两个迭代器。

public class ZigzagIterator {
    private Iterator<Integer> i, j, temp;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        i = v2.iterator();
        j = v1.iterator();
    }

    public int next() {
        if (j.hasNext()) {
            temp = i;
            i = j;
            j = temp;
        }
        return i.next();
    }

    public boolean hasNext() {
        return i.hasNext() || j.hasNext();
    }
}

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

 

这道题的followup是如果给K个一维的向量怎么处理。很显然如果有K个向量，那么自然还是需要K个迭代器来帮助输出来自于每个向量的元素，但是这里用temp来替换每个迭代器就不管用了。这里正确的思路是用一个queue存储所有的迭代器。每次调用next()函数的时候，就从queue中弹出一个迭代器。如果这个被弹出的迭代器里还有下一个元素，则弹出这个元素，并把这个迭代器再放回queue的末端；如果这个迭代器空了则无需再放回queue。

public class ZigzagIterator {
    private Queue<Iterator> queue;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        queue = new LinkedList<>();
        if (!v1.isEmpty()) {
            queue.offer(v1.iterator());
        }
        if (!v2.isEmpty()) {
            queue.offer(v2.iterator());
        }
    }

    public int next() {
        Iterator cur = queue.poll();
        Integer res = (Integer) cur.next();
        if (cur.hasNext()) {
            queue.offer(cur);
        }
        return res;
    }

    public boolean hasNext() {
        return !queue.isEmpty();
    }
}

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

 

LeetCode 题目总结