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

思路Coding化:

         i
vec = [[1,2], [3], [4] ]
        j

update pointer: 當j到了當前list的最後一位則update
next: return 當前的值，記得每次next，j都要加一
hasNext: 先update pointer(看j是否為最後一位)，接著return i是否超過vec長度


class Vector2D:

    def __init__(self, vec: List[List[int]]):
        self.vector = vec
        self.i = 0 
        self.j = 0 
    
    def update_pointers(self):
        while self.i < len(self.vector) and self.j == len(self.vector[self.i]):
            self.i += 1
            self.j = 0

    def next(self) -> int:
        if self.hasNext():
            result = self.vector[self.i][self.j]
            self.j += 1
            return result
        else:
            return "Exception Error"

    def hasNext(self) -> bool:
        self.update_pointers()
        return self.i < len(self.vector)

Time Complexity: O(N)

Space Complexity: O(1)