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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
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。
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。
1 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { 2 3 int carry=0; 4 struct ListNode* head = NULL; 5 struct ListNode* tail = NULL; 6 while(l1 || l2) 7 { 8 struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode)); 9 if(l1!= NULL && l2!= NULL) 10 { 11 new_node->val = (l1->val+l2->val+carry)%10; 12 carry = (l1->val+l2->val+carry)/10; 13 l1 = l1->next; 14 l2 = l2->next; 15 } 16 else if(l1 == NULL) 17 { 18 new_node->val = (l2->val+carry)%10; 19 carry = (l2->val+carry)/10; 20 l2 = l2->next; 21 } 22 else if(l2 == NULL) 23 { 24 new_node->val = (l1->val+carry)%10; 25 carry = (l1->val+carry)/10; 26 l1 = l1->next; 27 } 28 29 new_node->next = NULL; 30 if(head == NULL) 31 { 32 head = new_node; 33 tail = head; 34 } 35 else 36 { 37 tail->next = new_node; 38 tail = tail->next; 39 } 40 } 41 if(carry) 42 { 43 struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode)); 44 new_node->val = 1; 45 new_node->next = NULL; 46 tail->next = new_node; 47 tail = tail->next; 48 } 49 return head; 50 }
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原文地址：http://www.cnblogs.com/madking/p/4657449.html
