用链表表示的两个数相加

1 题目

You are giventwo linked lists representing two non-negative numbers. The digits are storedin reverse order and each of their nodes contain a single digit. Add the twonumbers and return it as a linked list.

Input: (2 -> 4 -> 3)+ (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

2 分析

该题目属于链表的相加，需要注意的有：

(1) 考虑两个链表的长度，尤其是链表为空时也能处理。

(2) 每个结点只表示一位数字。

(3) 当链表末尾结点相加后若有进位，则需要申请新的结点存储信息。

3 实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *    int val;
 *    ListNode *next;
 *    ListNode(int x) : val(x), next(NULL) {}
 * };
 */

intlistLength(ListNode* head)
{
      return head ? 1 + listLength(head ->next) : 0;
}

ListNode*Solution::addTwoNumbers(ListNode *l1, ListNode *l2)
{
      if (listLength(l1) < listLength(l2))
      {
             return addTwoNumbers(l2, l1);
      }
      ListNode *head1 = l1, *head2 = l2;
      int inc = 0;
      bool isEnd = false;
      while (head2)
      {
             int val = head1 -> val + head2-> val + inc;
             head1 -> val = val % 10;
             inc = val / 10;
             if (head1 -> next)
             {
                    head1 = head1 -> next;
             }
             else
             {
                    isEnd = true;
             }
             head2 = head2 -> next;
      }
      while (inc)
      {
             int val = isEnd ? inc : head1 ->val + inc;
             if (isEnd)
             {
                    head1 -> next = newListNode(val % 10);
             }
             else
             {
                    head1 -> val = val % 10;
             }
             inc = val / 10;
             if (head1 -> next)
             {
                    head1 = head1 -> next;
             }
             else
             {
                    isEnd = true;
             }
      }
      return l1;
}

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