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;

    }

}

View Code

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;