Given a nested list of integers, implement an iterator to flatten it.
Each element is either an integer, or a list -- whose elements may also be integers or other lists.
Example 1:
Given the list [[1,1],2,[1,1]],
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].
Example 2:
Given the list [1,[4,[6]]],
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].
Solution1:Stack
Time Complexity: O(N) Space Complexity: O(N)
Solution2:Stack Round1
Time Complexity: O(N) Space Complexity: O(N)
Solution1 Code:
public class NestedIterator implements Iterator<Integer> {
Deque<NestedInteger> stack = new ArrayDeque<>();
public NestedIterator(List<NestedInteger> nestedList) {
for(int i = nestedList.size() - 1; i >= 0; i--) {
stack.push(nestedList.get(i));
}
}
@Override
public Integer next() {
return stack.pop().getInteger();
}
@Override
public boolean hasNext() {
while(!stack.isEmpty()) {
NestedInteger curr = stack.peek();
if(curr.isInteger()) {
return true;
}
stack.pop();
for(int i = curr.getList().size() - 1; i >= 0; i--) {
stack.push(curr.getList().get(i));
}
}
return false;
}
}
Solution2 Round1 Code:
public class NestedIterator implements Iterator<Integer> {
private Deque<NestedInteger> stack;
public NestedIterator(List<NestedInteger> nestedList) {
stack = new ArrayDeque<>();
for(int i = nestedList.size() - 1; i >= 0; i--) {
stack.push(nestedList.get(i));
}
}
@Override
public Integer next() {
return stack.pop().getInteger();
}
@Override
public boolean hasNext() {
while(stack.size() > 0) {
NestedInteger elem = stack.peek();
if(elem.isInteger()) {
return true;
}
else {
elem = stack.pop();
List<NestedInteger> list = elem.getList();
for(int i = list.size() - 1; i >= 0; i--) {
stack.push(list.get(i));
}
}
}
return false;
}
}