590. N-ary Tree Postorder Traversal

/*
// Definition for a Node.
class Node {
    public int val;
    public List children;

    public Node() {}

    public Node(int _val,List _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
   	public List postorder(Node root) {
		List out = new ArrayList();
		postorder(root, out);
		return out;
	}

	private void postorder(Node root, List out) {
		if( null != root){
		for (int i = 0; i < root.children.size(); i++)
			postorder(root.children.get(i), out);
		out.add(root.val);
        }
	}
}

788. Rotated Digits

class Solution {
 	public static int rotatedDigits(int N) {

		int good = N;
		for (int i = 1; i <= N; i++) {
			boolean valid = true;
			int current = i;
			int converted = i;
			inner: {
				int m = 1;
				while (current != 0) {
					int digit = current % 10; // get current digit

					// Convert digits regard to the specific rule
					if (digit == 2) {
						converted -= (digit * m);
						converted += (5 * m);
					} else if (digit == 5) {
						converted -= (digit * m);
						converted += (2 * m);
					} else if (digit == 6) {
						converted -= (digit * m);
						converted += (9 * m);
					} else if (digit == 9) {
						converted -= (digit * m);
						converted += (6 * m);
					}

					if (digit == 3 || digit == 4 || digit == 7) {
						valid = false;
						good--;
						break inner;
					}
					current /= 10;
					m *= 10;// next digit
				}

			}
			if (converted == i && valid)
				good--;
		}
		return good;
	}
}

860. Lemonade Change

class Solution {
		public static boolean lemonadeChange(int[] bills) {
		HashMap d = new HashMap();
		d.put(5, 0);
		d.put(10, 0);
		for (int i = 0; i < bills.length; i++) {
			if (bills[i] == 5) {
				d.put(5, d.get(5) + 1);
			} else if (bills[i] == 10) {
				if (d.get(5) <= 0)
					return false;
				d.put(5, d.get(5) - 1);
				d.put(10, d.get(10) + 1);
			} else if (bills[i] == 20) {
				if ((d.get(10) - 1) < 0) { // no 10's
					if (d.get(5) - 3 < 0)
						return false;
				} else {// there are 10's
					if (d.get(5) - 1 = 1 && d.get(5) >= 1) {
					d.put(5, d.get(5) - 1);
					d.put(10, d.get(10) - 1);
				} else if (d.get(5) - 3 >= 0) {
					d.put(5, d.get(5) - 3);
				}
			}
		}
		return true;
	}
}

Multiples of 3 and 5

Answer is 233168 pragmatically.

I did not use the Math to get the answer.

public class Main {
public static int multiple3and5below1000() { 
int sum = 0; 
for (int i = 3; i < 1000; i++) { 
if ((i >= 3 && ((i % 3) == 0) || (i >= 5 && ((i % 5) == 0)))) 
sum += i; } 
return sum; }
 public static void main(String[] args) { 
 System.out.println(multiple3and5below1000()); }
}

448. Find All Numbers Disappeared in an Array

class Solution {
  public List findDisappearedNumbers(int[] nums) {
	        boolean[] data = new boolean [nums.length];
	        for ( int i = 0 ; i < nums.length ; i++) {
	        	data[nums[i] - 1 ] = true;
	        }
	        List output = new ArrayList();
	        for ( int i = 0 ; i < data.length ; i++) {
	        	if ( ! data[i] )output.add(i+1);
	        }
	        return output;
	    }
}