LeetCode----- 2.Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

給出了兩個非空鏈表，表示兩個非負整數。數字以倒序存儲，每個節點都包含一個數字。添加兩個數字並將其作爲一個鏈表返回。

提示：本題解法是將兩個鏈表對應的節點進行求和，類似與數學中的2個數相加，注意一點別忽略進位。

public class AddTwoNumbers {

	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		ListNode result = new ListNode(0);
		if(l1 ==null && l2 == null) {
			return result.next;
		}		
		ListNode currentNode = result;
		int sum = 0;
		while(l1 != null || l2 != null || sum != 0) {
			if(l1 != null) {
				sum += l1.val;
				l1 = l1.next;
			}
			if(l2 != null) {
				sum += l2.val;
				l2 = l2.next;
			}
			ListNode node = new ListNode(sum%10);
			currentNode.next = node;
			currentNode = node;
			sum = sum/10;
		}
		return result.next;
	} 
	
	public static void main(String[] args) {
		ListNode l10 = new ListNode(2);
		ListNode l11 = new ListNode(4);
		ListNode l12 = new ListNode(3);
		l10.next = l11;
		l11.next = l12;
		l12.next = null;
		
		ListNode l20 = new ListNode(5);
		ListNode l21 = new ListNode(6);
		ListNode l22 = new ListNode(4);
		l20.next = l21;
		l21.next = l22;
		l22.next = null;
		
		ListNode node = addTwoNumbers(l10, l20);
		while(node != null) {
			if(node.next == null) {
				System.out.println(node.val);
			}else {
				System.out.print(node.val+"->");
			}
			node = node.next;
		}
	}
}

class ListNode {
	int val;
	ListNode next;
	ListNode(int x) { 
		val = x;
	}
}