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
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { 9 struct ListNode *phead; 10 struct ListNode *cur1; 11 struct ListNode *cur2; 12 struct ListNode *curp; 13 struct ListNode *dele; 14 struct ListNode *pnew; 15 phead = NULL; 16 cur1 = l1; 17 cur2 = l2; 18 int flag = 0; 19 while(cur1 != NULL || cur2 != NULL) 20 { 21 if(phead == NULL) 22 { 23 curp = (struct ListNode *)malloc(sizeof(struct ListNode)); 24 memset(curp,0,sizeof(struct ListNode)); 25 phead = curp; 26 } 27 else 28 { 29 pnew = (struct ListNode *)malloc(sizeof(struct ListNode)); 30 memset(pnew,0,sizeof(struct ListNode)); 31 curp->next = pnew; 32 curp = pnew; 33 } 34 if(cur1 == NULL) 35 { 36 curp->val = (cur2->val + flag) % 10; 37 flag = (cur2->val + flag) / 10; 38 dele = cur2; 39 cur2 = cur2->next; 40 free(dele); 41 dele = NULL; 42 } 43 else if(cur2 == NULL) 44 { 45 curp->val = (cur1->val + flag) % 10; 46 flag = (cur1->val + flag) / 10; 47 dele = cur1; 48 cur1 = cur1->next; 49 free(dele); 50 dele = NULL; 51 } 52 else 53 { 54 curp->val = (cur1->val + flag + cur2->val) % 10; 55 flag = (cur1->val + flag + cur2->val) /10; 56 dele = cur2; 57 cur2 = cur2->next; 58 free(dele); 59 dele = NULL; 60 dele = cur1; 61 cur1 = cur1->next; 62 free(dele); 63 dele = NULL; 64 } 65 66 } 67 if(flag == 1) //最后要注意进位的问题！！！ 68 { 69 pnew = (struct ListNode *)malloc(sizeof(struct ListNode)); 70 memset(pnew,0,sizeof(struct ListNode)); 71 curp->next = pnew; 72 pnew->val = 1; 73 return phead; 74 } 75 return phead; 76 }