原题链接在这里：https://leetcode.com/problems/flatten-2d-vector/
题目：
Implement an iterator to flatten a 2d vector.
Example:
Input: 2d vector =
[
[1,2],
[3],
[4,5,6]
]
Output:[1,2,3,4,5,6]
Explanation: By calling next repeatedly until hasNext returns false,
the order of elements returned by next should be:[1,2,3,4,5,6].
Follow up:
As an added challenge, try to code it using only iterators in C++ or iterators in Java.
题解：
用两个index 分别记录list 的 index 和当前 list的element index.
Time Complexity: Vector2D() O(1). hasNext() O(vec2d.size()). next() O(1). Space: O(1).
AC Java:
public class Vector2D {
List<List<Integer>> listOfList;
int listIndex;
int elemIndex;
public Vector2D(List<List<Integer>> vec2d) {
listOfList = vec2d;
listIndex = 0;
elemIndex = 0;
}
public int next() {
return listOfList.get(listIndex).get(elemIndex++);
}
public boolean hasNext() {
while(listIndex < listOfList.size()){
if(elemIndex < listOfList.get(listIndex).size()){
return true;
}else{
listIndex++;
elemIndex = 0;
}
}
return false;
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* Vector2D i = new Vector2D(vec2d);
* while (i.hasNext()) v[f()] = i.next();
*/
Follow up 要用Iterator class.
Time Complexity: Vector2D() O(1). hasNext() O(vec2d.size()). next() O(1). Space: O(1).
AC Java:
public class Vector2D implements Iterator<Integer> {
Iterator<List<Integer>> i;
Iterator<Integer> j;
public Vector2D(List<List<Integer>> vec2d) {
i = vec2d.iterator();
}
@Override
public Integer next() {
return j.next();
}
@Override
public boolean hasNext() {
while((j==null || !j.hasNext()) && i.hasNext()){
j = i.next().iterator();
}
return j!=null && j.hasNext();
}
}
/**
* Your Vector2D object will be instantiated and called as such:
* Vector2D i = new Vector2D(vec2d);
* while (i.hasNext()) v[f()] = i.next();
*/