README_EN.md

251. Flatten 2D Vector

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Description

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.

Solutions

Python3

class Vector2D:
    def __init__(self, vec: List[List[int]]):
        self.flatten = []
        for item in vec:
            for e in item:
                self.flatten.append(e)
        self.cur = -1

    def next(self) -> int:
        self.cur += 1
        return self.flatten[self.cur]

    def hasNext(self) -> bool:
        return self.cur < len(self.flatten) - 1


# Your Vector2D object will be instantiated and called as such:
# obj = Vector2D(vec)
# param_1 = obj.next()
# param_2 = obj.hasNext()

Java

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