给出链表，求链表上面数据之和代码,Java代码库,德仔网

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【Java】给出链表，求链表上面数据之和

作者:会飞的章鱼 / 发布于2015/6/2/ 578

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


	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
	 if(!l1){
	 return NULL;
	 }
	 // ListNode *curr1=l1;
	 // ListNode*next1= NULL;
	 // ListNode*pre1 = NULL;
	 // while(curr1){
	 // next1 = curr1->next;
	 // curr1->next = pre1;
	 // pre1 = curr1;
	 // curr1 = next1;
	 // }
	 // // get pre1 return pre;
	 if(!l2){
	 return NULL;
	 }
	 // ListNode *curr2=l2;
	 // ListNode*next2= NULL;
	 // ListNode*pre2 = NULL;
	 // while(curr2){
	 // next2 = curr2->next;
	 // curr2->next = pre2;
	 // pre2 = curr2;
	 // curr2 = next2;
	 // }
	 //get pre2;
	 ListNode*pre1 = l1;
	 ListNode*pre2 = l2;
	 ListNode * result = new ListNode(0);
	 ListNode * sp = result;
	 int tmp =0;
	 while(pre1!=NULL || pre2 !=NULL){
	 int k1;
	 int k2;
	 if(pre1==NULL){
	 k1 =0;
	 }else{
	 k1 = pre1->val;
	 }
	
	 if(pre2==NULL){
	 k2 =0;
	 }else{
	 k2 = pre2->val;
	 }
	 result->val = (tmp+k1+k2)%10;
	 tmp = (tmp+k1+k2)/10;
	 if((pre1!=NULL &&pre1->next!=NULL) || (pre2!=NULL &&pre2->next !=NULL)){
	 result->next = new ListNode(0);
	 }else{
	 result->next = NULL;
	 break;
	 }
	 result = result->next;
	 if(pre1!=NULL && pre1->next !=NULL){
	 pre1 = pre1->next;
	
	 }else{
	 pre1 = NULL;
	 }
	 if(pre2!=NULL && pre2->next !=NULL){
	 pre2 = pre2->next;
	 }else{
	 pre2 =NULL;
	 }
	
	 }
	 if(tmp>0){
	 result->next = new ListNode(0);
	 result = result->next;
	 result->val = tmp;
	 result->next =NULL;
	 }
	 return sp;
	 }

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