LeetCode.002 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

题意

给你两个链表，求出这两个链表所代表数字的和。

链表2 -> 4 -> 3 代表数字 324，链表5 -> 6 -> 4代表数字465，求出和的结果也用链表倒序表示。

思路

1.两个链表都从头节点开始依次相加，相加节点的个位数保存在新节点中，进位保存在carry中。

2.如果某一个链表遍历完了，则另一个链表和carry接着相加，直到遍历完。

3.要注意的是最后的进位（carry）如果等于1，则还需要新建立个节点存储此进位，再添加到结果链表的尾部就可以了。

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    
     int carry=0;
     struct ListNode* head = NULL;
     struct ListNode* tail = NULL;
     while(l1 || l2)
     {
         struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode));
         if(l1!= NULL && l2!= NULL)
         {
             new_node->val = (l1->val+l2->val+carry)%10;
             carry = (l1->val+l2->val+carry)/10;
             l1 = l1->next;
             l2 = l2->next;
         }
         else if(l1 == NULL)
         {
             new_node->val = (l2->val+carry)%10;
             carry = (l2->val+carry)/10;
             l2 = l2->next;
         }
         else if(l2 == NULL)
         {
             new_node->val = (l1->val+carry)%10;
             carry = (l1->val+carry)/10;
             l1 = l1->next;
         }
        
         new_node->next = NULL;
         if(head == NULL)
         {
             head = new_node;
             tail = head;
         }
         else
         {
             tail->next = new_node;
             tail  = tail->next;
         }
     }
    if(carry)
    {
        struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode));
        new_node->val = 1;
        new_node->next = NULL;
        tail->next = new_node;
        tail  = tail->next;
    }
    return head;
}