[LeetCode] 251. Flatten 2D Vector

Design an iterator to flatten a 2D vector. It should support the next and hasNext operations.

Implement the Vector2D class:

• Vector2D(int[][] vec) initializes the object with the 2D vector vec.
• next() returns the next element from the 2D vector and moves the pointer one step forward. You may assume that all the calls to next are valid.
• hasNext() returns true if there are still some elements in the vector, and false otherwise.

Example 1:

Input
["Vector2D", "next", "next", "next", "hasNext", "hasNext", "next", "hasNext"]
[[[[1, 2], [3], [4]]], [], [], [], [], [], [], []]
Output
[null, 1, 2, 3, true, true, 4, false]

Explanation
Vector2D vector2D = new Vector2D([[1, 2], [3], [4]]);
vector2D.next();    // return 1
vector2D.next();    // return 2
vector2D.next();    // return 3
vector2D.hasNext(); // return True
vector2D.hasNext(); // return True
vector2D.next();    // return 4
vector2D.hasNext(); // return False

Constraints:

• 0 <= vec.length <= 200
• 0 <= vec[i].length <= 500
• -500 <= vec[i][j] <= 500
• At most 105 calls will be made to next and hasNext.

Follow up: As an added challenge, try to code it using only iterators in C++ or iterators in Java.

展开二维向量。

这也是一道考察 iterator 的题目，只不过是把 iterate 的对象变成了一个二维数组。比较偷懒的办法就是先把所有元素放入一个队列，然后遍历队列即可。

Java实现

class Vector2D {
    private Queue<Integer> queue = new LinkedList<>();

    public Vector2D(int[][] v) {
        for (int i = 0; i < v.length; i++) {
            for (int j = 0; j < v[i].length; j++) {
                queue.offer(v[i][j]);
            }
        }
    }

    public int next() {
        return queue.poll();
    }

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

/**
 * Your Vector2D object will be instantiated and called as such:
 * Vector2D obj = new Vector2D(v);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

 

不使用队列也能做，无非考察的是你对遍历二维矩阵的坐标的敏感程度。注意这道题给的二维向量，每一行的元素个数并不一样，所以不能用类似遍历二维矩阵那样的方式去遍历。

Java实现

class Vector2D {
    int[][] v;
    int i = 0;
    int j = 0;

    public Vector2D(int[][] v) {
        this.v = v;
    }

    public int next() {
        if (hasNext()) {
            return v[i][j++];
        } else {
            return -1;
        }
    }

    public boolean hasNext() {
        while (i < v.length && j == v[i].length) {
            i++;
            j = 0;
        }
        return i < v.length;
    }
}

/**
 * Your Vector2D object will be instantiated and called as such:
 * Vector2D obj = new Vector2D(v);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

 

LeetCode 题目总结