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 <= 2000 <= 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
Solution 1
PythonJavaC++GoTypeScript
class Vector2D:
def __init__(self, vec: List[List[int]]):
self.i = 0
self.j = 0
self.vec = vec
def next(self) -> int:
self.forward()
ans = self.vec[self.i][self.j]
self.j += 1
return ans
def hasNext(self) -> bool:
self.forward()
return self.i < len(self.vec)
def forward(self):
while self.i < len(self.vec) and self.j >= len(self.vec[self.i]):
self.i += 1
self.j = 0
# Your Vector2D object will be instantiated and called as such:
# obj = Vector2D(vec)
# param_1 = obj.next()
# param_2 = obj.hasNext()(code-box)
class Vector2D {
private int i;
private int j;
private int[][] vec;
public Vector2D(int[][] vec) {
this.vec = vec;
}
public int next() {
forward();
return vec[i][j++];
}
public boolean hasNext() {
forward();
return i < vec.length;
}
private void forward() {
while (i < vec.length && j >= vec[i].length) {
++i;
j = 0;
}
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* Vector2D obj = new Vector2D(vec);
* int param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/(code-box)
class Vector2D {
public:
Vector2D(vector<vector<int>>& vec) {
this->vec = move(vec);
}
int next() {
forward();
return vec[i][j++];
}
bool hasNext() {
forward();
return i < vec.size();
}
private:
int i = 0;
int j = 0;
vector<vector<int>> vec;
void forward() {
while (i < vec.size() && j >= vec[i].size()) {
++i;
j = 0;
}
}
};
/**
* Your Vector2D object will be instantiated and called as such:
* Vector2D* obj = new Vector2D(vec);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/(code-box)
type Vector2D struct {
i, j int
vec [][]int
}
func Constructor(vec [][]int) Vector2D {
return Vector2D{vec: vec}
}
func (this *Vector2D) Next() int {
this.forward()
ans := this.vec[this.i][this.j]
this.j++
return ans
}
func (this *Vector2D) HasNext() bool {
this.forward()
return this.i < len(this.vec)
}
func (this *Vector2D) forward() {
for this.i < len(this.vec) && this.j >= len(this.vec[this.i]) {
this.i++
this.j = 0
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* obj := Constructor(vec);
* param_1 := obj.Next();
* param_2 := obj.HasNext();
*/(code-box)
class Vector2D {
i: number;
j: number;
vec: number[][];
constructor(vec: number[][]) {
this.i = 0;
this.j = 0;
this.vec = vec;
}
next(): number {
this.forward();
return this.vec[this.i][this.j++];
}
hasNext(): boolean {
this.forward();
return this.i < this.vec.length;
}
forward(): void {
while (this.i < this.vec.length && this.j >= this.vec[this.i].length) {
++this.i;
this.j = 0;
}
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* var obj = new Vector2D(vec)
* var param_1 = obj.next()
* var param_2 = obj.hasNext()
*/(code-box)
