LeetCode – Flatten Nested List Iterator (Java)

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.

For example, 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].

Java Solution 1

public class NestedIterator implements Iterator<Integer> {
    Stack<NestedInteger> stack = new Stack<NestedInteger>();
 
    public NestedIterator(List<NestedInteger> nestedList) {
        if(nestedList==null)
            return;
 
        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 top = stack.peek();
            if(top.isInteger()){
                return true;
            }else{
                stack.pop();
                for(int i=top.getList().size()-1; i>=0; i--){
                    stack.push(top.getList().get(i));
                }
            }
        }
 
        return false;
    }
}

Java Solution 2

public class NestedIterator implements Iterator<Integer> {
    Stack<Iterator<NestedInteger>> stack = new Stack<Iterator<NestedInteger>>();
    Integer current;
 
    public NestedIterator(List<NestedInteger> nestedList) {
        if(nestedList==null)
            return;
 
        stack.push(nestedList.iterator());    
    }
 
    @Override
    public Integer next() {
        Integer result = current;
        current = null;
        return result;
    }
 
    @Override
    public boolean hasNext() {
        while(!stack.isEmpty() && current==null){
            Iterator<NestedInteger> top = stack.peek();
            if(!top.hasNext()){
                stack.pop();
                continue;
            }
 
            NestedInteger n = top.next();
            if(n.isInteger()){
                current = n.getInteger();
                return true;
            }else{
                stack.push(n.getList().iterator());
            }
        }
 
        return false;
    }
}
Category >> Algorithms  
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  • aman

    Awsome . Thx