【leetcode with java】2 Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

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

Tags：Linked List Math

</pre><p><strong>思路（模擬手算）：</strong>初始進位數carry爲0，用兩個指針p1、p2分別指向兩個鏈表的第一個節點。然後計算當前位的數值：(p1.val+p2.val+carry)%10和進位數（p1.val+p2.val+carry）/10。  然後p1和p2同時向下走一步，進行同樣計算。</p><p>需要注意的是，兩個鏈表可能不一樣長。</p><p></p><p>上碼：</p><p><pre name="code" class="java">/*
 * O(n)
 */
public class Solution {
	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
	       ListNode sum;
	       int carry = 0;//進位 
	       
	       ListNode p1=l1,p2=l2;
	       int s = p1.val + p2.val + carry;
	       carry = s/10;
	       sum = new ListNode(s%10);
	       ListNode p=sum;
	       p1 = p1.next;
	       p2 = p2.next;
	       
	       while(p1!=null && p2!=null){
	    	   s = p1.val + p2.val + carry;
	    	   carry = s/10;
	    	   ListNode node = new ListNode(s%10);
	    	   p.next = node;
	    	   p = p.next;
	    	   p1 = p1.next;
	    	   p2 = p2.next;
	       }

	       if (p1!=null || p2!=null) {
	    	   p1 = p1==null? p2:p1;
	    	   while(p1!=null){
				s = p1.val+carry;
				carry = s/10;
				ListNode node = new ListNode(s%10);
				p.next = node;
				p = p.next;
				p1 = p1.next;
	    	   }
	       }
	       if (carry > 0) {
			ListNode node = new ListNode(carry);
			p.next = node;
	       }
	       return sum;
	}
	
	
}