题2、Add Two Numbers

题目

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

思路

这道题就是在考链表，啥乱七八糟的关系搞清楚就行了，我也是懵了半天才反应过来。
另外需要注意的就是进位别把进位给搞丢了。
还有就是别忘了，给出的两个数可不一定等长，注意处理各种特殊情况，不过也不慌，坑踩多了就记住了。

代码

本人菜鸡，大佬们想看就看看，不想看也就那样

public class Add_Two_Numbers {

	public static void main(String[] args) {
		
		ListNode a = new ListNode( 2 );
		ListNode b = new ListNode( 4 );
		ListNode c = new ListNode( 3 );
		
		ListNode d = new ListNode( 5 );
		ListNode e = new ListNode( 6 );
		ListNode f = new ListNode( 7 );
		
		a.next = b;
		b.next = c;
		
		d.next = e;
		e.next = f;
		
		ListNode o = addTwoNumbers( a, d );

		while( o != null ) {
			System.out.print( o.val );
			o = o.next;
		}
			
	}
	
	 public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
	        ListNode first = new ListNode(0);
	    	ListNode before = first;
	    	
	    	int up = 0;
	    	
	    	while( l1!=null || l2!=null || up!=0) {
	    		if( l1!=null && l2!=null ) {
	    		    before.next = new ListNode( (l1.val + l2.val + up)%10 );
	    		    up = (l1.val + l2.val + up)/10;
	    		    l1 = l1.next;
	    		    l2 = l2.next;
	    		}else if( l1==null && l2!=null ) {
	    			before.next = new ListNode( (l2.val + up)%10 );
	    		    up = (l2.val + up)/10;
	    		    l2 = l2.next;
	    		}else if( l1!=null && l2==null ) {
	    			before.next = new ListNode( (l1.val + up)%10 );
	    		    up = (l1.val + up)/10;
	    		    l1 = l1.next;
	    		}else if( l1==null && l2==null && up != 0 ) {
	    			before.next = new ListNode( up );
	    			up = 0;
	    		}
	    		
	    		before = before.next;
	    		
	    	}
	    	
			return first.next;
	        
	    }
}
//节点
class ListNode {
	int val;
	ListNode next;
	ListNode(int x) { val = x; }
  }