Add Two Numbers

Add Two Numbers

问题：

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

思路：

　　归并排序 最后面的操作而已

我的代码：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null) return l2;
        if(l2 == null)  return l1;
        int plus = 0;
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while(l1 != null && l2 != null)
        {
            int val = plus + l1.val + l2.val;
            if(val >= 10)
            {
                ListNode node = new ListNode(val-10);
                cur.next = node;
                plus = 1;
            }
            else
            {
                ListNode node = new ListNode(val);
                cur.next = node;
                plus = 0;
            }
            l1 = l1.next;
            l2 = l2.next;
            cur = cur.next;
        }
        while(l1 != null)
        {
            int val = plus + l1.val;
            if(val >= 10)
            {
                ListNode node = new ListNode(val-10);
                cur.next = node;
                plus = 1;
            }
            else
            {
                ListNode node = new ListNode(val);
                cur.next = node;
                plus = 0;
            }
            l1 = l1.next;
            cur = cur.next;
        }
        while(l2 != null)
        {
            int val = plus + l2.val;
            if(val >= 10)
            {
                ListNode node = new ListNode(val-10);
                cur.next = node;
                plus = 1;
            }
            else
            {
                ListNode node = new ListNode(val);
                cur.next = node;
                plus = 0;
            }
            l2 = l2.next;
            cur = cur.next;
        }
        if(plus != 0)
        {
            ListNode node = new ListNode(1);
            cur.next = node;
        }
        return dummy.next;
    }
}

学习之处：

• 思路很简单，不过有两个地方可以进行简化，使代码看起来更加整洁 carry的获得方法，可以改成如下所示：

  int sum = carry + l1.val + l2.val;
  point.next = new ListNode(sum % 10);
  carry = sum / 10;