大概意思就是把2D list里面的element全部遍歷一遍。
注意啊偎球，一開始理解題意搞錯：我以為是必須要排序正確洒扎，所以上來就PriorityQueue+HashMap搞得無比復(fù)雜辑甜。其實，這個跟一個nxn的matrix遍歷袍冷，是沒區(qū)別的拉磷醋。
所有來個x,y，把2d list跑一變胡诗。
/*
Implement an iterator to flatten a 2d vector.
For example,
Given 2d vector =
[
[1,2],
[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:
How many variables do you need to keep track?
Two variables is all you need. Try with x and y.
Beware of empty rows. It could be the first few rows.
To write correct code, think about the invariant to maintain. What is it?
The invariant is x and y must always point to a valid point in the 2d vector. Should you maintain your invariant ahead of time or right when you need it?
Not sure? Think about how you would implement hasNext(). Which is more complex?
Common logic in two different places should be refactored into a common method.
Tags: Design
Similar Problems: (M) Binary Search Tree Iterator, (M) Zigzag Iterator, (M) Peeking Iterator
*/
/*
Thoughts:
As hint indicates: use 2 pointers to hold position.
Use hasNext to validate (x,y) and move x.
Use next() to return (x,y) and move it(regardless of correctness, which is determined by hasNext())
*/
public class Vector2D {
private int x;
private int y;
private List<List<Integer>> list;
public Vector2D(List<List<Integer>> vec2d) {
if (vec2d == null) {
return;
}
this.x = 0;
this.y = 0;
this.list = vec2d;
}
public int next() {
int rst = list.get(x).get(y);
if (y + 1 >= list.get(x).size()) {
y = 0;
x++;
} else {
y++;
}
return rst;
}
public boolean hasNext() {
if (list == null) {
return false;
}
while (x < list.size() && list.get(x).size() == 0) {
x++;
y = 0;
}
if (x >= list.size()) {
return false;
}
if (y >= list.get(x).size()) {
return false;
}
return true;
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* Vector2D i = new Vector2D(vec2d);
* while (i.hasNext()) v[f()] = i.next();
*/
