LeetCode Algorithm 02_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

分析：逐位相加，考虑进位。

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
         if (!l1 || !l2) {
             return NULL;
         }
         int carry = ;  //进位
         ListNode * result = new ListNode();
         ListNode * first = result;  //追踪结果链表的当前节点
         ListNode * pre = NULL;  //追踪结果链表的前一个节点
         //当l1和l2均没有超过链表末尾节点时
         while (l1 && l2) {
             first->val = (carry + l1->val + l2->val) % ;
             carry = (carry + l1->val + l2->val) / ;
             if (pre == NULL)
                 pre = first;
             else {
                 pre->next = first;
                 pre = first;
             }
             first = new ListNode();
             l1 = l1->next;
             l2 = l2->next;
         }
         //当l1和l2都超过链表末尾节点时
         if (!l1 && !l2) {
             if (carry == ) {
                 first->val = carry;
                 pre->next = first;
             }
             return result;
         }
         //当l1超过末尾而l2尚未超过时
         if (!l1 && l2) {
             while (l2) {
                 first->val = (carry + l2->val) % ;
                 carry = (carry + l2->val) / ;
                 if (pre == NULL)
                     pre = first;
                 else {
                     pre->next = first;
                     pre = first;
                 }
                 first = new ListNode();
                 l2 = l2->next;
             }
             if (carry == ) {
                 first->val = ;
                 pre->next = first;
             }
             return result;
         }
         //当l2超过末尾而l1尚未超过时
         if (!l2 && l1) {
             while (l1) {
                 first->val = (carry + l1->val) % ;
                 carry = (carry + l1->val) / ;
                 if (pre == NULL)
                     pre = first;
                 else {
                     pre->next = first;
                     pre = first;
                 }
                 first = new ListNode();
                 l1 = l1->next;
             }
             if (carry == ) {
                 first->val = ;
                 pre->next = first;
             }
             return result;
         }
 
     }
 };