251. Flatten 2D Vector

Implement a iterator to flatten a 2d vector.

For example,
Given 2d vector =

[  [up],  [3],  [4,5,6]]

By calling next repeatedly until hasnext returns FALSE, the order of elements returned by next should be: [1,2,3,4,5,6] .

Hint:

2. How many variables does you need to keep track?
4. Variables are all need. Try with x and y .
6. Beware of empty rows. It could be the first few rows.
8. To write correct code, think on the invariant to maintain. What is it?
10. The invariant is x and y must all-point-a valid point in the 2d vector. Should maintain your invariant ahead of time or right when do you need it?
12. Not sure? Think about what you would implement hasNext() . Which is more complex?
14. Common logic in the different places should is refactored into a Common method.

 Public classvector2d { Publicilist<ilist<int>> v {Get;Set;} Private intVerticalcount {Get;Set;} Private intIndex {Get;Set;}  PublicVector2d (ilist<ilist<int>>vec2d) {v=vec2d; Verticalcount=0; Index=0; }     Public BOOLHasnext () {if(Verticalcount >= v.count ())return false; if(Verticalcount = = V.count ()-1&& index >= V[verticalcount]. Count ())return false; if(Index >=V[verticalcount]. Count ()) {Verticalcount++;  while(Verticalcount < V.count () && V[verticalcount]. Count () = =0) {Verticalcount++; }            if(Verticalcount = = V.count ())return false; Else{Index=0; return true; }                   }        return true; }     Public intNext () {returnv[verticalcount][index++]; }}/** * Your vector2d'll be a called like this : * vector2d i = new vector2d (vec2d), * while (I.hasnext ()) v[f ()] = I.next ( ); */

251. Flatten 2D Vector