LeetCode——Add Two Numbers

public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		ListNode l1_r = l1;
		ListNode l2_r = l2;

		ListNode result = null;// 作为返回值
		// 处理第一个结点
		int add = 0;// 进位
		int s = l1_r.val + l2_r.val + add;
		result = new ListNode(s % 10);
		add = s / 10;
		l1_r = l1_r.next;
		l2_r = l2_r.next;
		//上面的result作为头结点的指针
		ListNode l3_r = result;//l3_r作为链表的指针遍历
		while (l1_r != null || l2_r != null) {
			// 同时都不为null
			ListNode l3 = null;
			int sum = 0;
			if (l1_r != null && l2_r != null) {
				sum = l1_r.val + l2_r.val + add;
				l3 = new ListNode(sum % 10);
				add = sum / 10;
				// 后移
				l1_r = l1_r.next;
				l2_r = l2_r.next;
			} else if (l1_r != null && l2_r == null) {
				sum = l1_r.val + add;
				l3 = new ListNode(sum % 10);
				add = sum / 10;
				// 后移
				l1_r = l1_r.next;
			} else {
				sum = l2_r.val + add;
				l3 = new ListNode(sum % 10);
				add = sum / 10;
				// 后移
				l2_r = l2_r.next;
			}
			l3_r.next = l3;
			l3_r = l3_r.next;
		}
		//判断最后的add
		/**
		 * 最后会有一种情况就是链表的计算都结束了，但是add中还有进位，此时要注意的是计算进位
		 */
		if (add != 0){
			ListNode last = new ListNode(add);
			l3_r.next = last;
		}
		return result;
	}