221. Add Two Numbers II【medium】

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 

Example

Given 6->1->7 + 2->9->5. That is, 617 + 295.

Return 9->1->2. That is, 912.

解法一：

 class Solution {
 public:
     /**
      * @param l1: the first list
      * @param l2: the second list
      * @return: the sum list of l1 and l2
      */
     ListNode *addLists2(ListNode *l1, ListNode *l2) {
         reverse(l1);
         reverse(l2);
         ListNode *dummy = new ListNode();
         ListNode *tail = dummy;
         int carry = ;

         while (l1 != NULL || l2 != NULL) {
             int sum = carry;
             if (l1 != NULL) {
                 sum += l1->val;
                 l1 = l1->next;
             }
             if (l2 != NULL) {
                 sum += l2->val;
                 l2 = l2->next;
             }
             if (sum > ) {
                 carry = ;
                 sum -= ;
             } else {
                 carry = ;
             }
             tail->next = new ListNode(sum);
             tail = tail->next;
         }

         if (carry == ) {
             tail->next = new ListNode();
             tail = tail->next;
         }

         reverse(dummy->next);

         return dummy->next;
     }

     void reverse(ListNode *&head) {
         ListNode *prev = NULL;

         while (head != NULL) {
             ListNode *temp = head->next;
             head->next = prev;
             prev = head;
             head = temp;
         }

         head = prev;
     }
 };

链表先翻转，再求和，然后再翻转

解法二：

 public class Solution {
     /**
      * @param l1: the first list
      * @param l2: the second list
      * @return: the sum list of l1 and l2
      */
     public ListNode addLists2(ListNode l1, ListNode l2) {
         Stack<Integer> temp1 = reverseNode(l1);
         Stack<Integer> temp2 = reverseNode(l2);  

         ListNode point = new ListNode(0);
         int flag = 0;  

         while((!temp1.isEmpty()) && (!temp2.isEmpty())){
             int value = temp1.pop() + temp2.pop() + flag;
             flag = value / 10;
             value = value % 10;
             ListNode cur = new ListNode(value);
             cur.next = point.next;
             point.next = cur;
         }  

         while(!temp1.isEmpty()){
             int value = temp1.pop() + flag;
             flag = value / 10;
             value = value % 10;
             ListNode cur = new ListNode(value);
             cur.next = point.next;
             point.next = cur;
         }  

         while(!temp2.isEmpty()){
             int value = temp2.pop() + flag;
             flag = value / 10;
             value = value % 10;
             ListNode cur = new ListNode(value);
             cur.next = point.next;
             point.next = cur;
         }  

         if(flag == 1){
             ListNode cur = new ListNode(1);
             cur.next = point.next;
             point.next = cur;
         }  

         return point.next;
     }  

     public Stack<Integer> reverseNode(ListNode temp){
         Stack<Integer> record = new Stack<Integer>();  

         while(temp != null){
             record.push(temp.val);
             temp = temp.next;
         }  

         return record;
     }
 }  

利用栈的先进后出，记录两个链表的节点值。然后计算对应位置的两个数字之和，如果有进位，赋值给flag并带入下一个计算。

参考@sunday0904 的代码