2. Add Two Numbers【medium】

2. Add Two Numbers【medium】

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解法一：

 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
         ListNode * dummy = new ListNode(INT_MIN);
         ListNode * head = dummy;
         int carry = ;

         while (l1 != NULL && l2 != NULL) {
             int sum = l1->val + l2->val + carry;
             carry = sum / ;
             head->next = new ListNode(sum % );
             head = head->next;
             l1 = l1->next;
             l2 = l2->next;
         }

         while (l1 != NULL) {
             int sum = l1->val + carry;
             carry = sum / ;
             head->next = new ListNode(sum % );
             head = head->next;
             l1 = l1->next;
         }

         while (l2 != NULL) {
             int sum = l2->val + carry;
             carry = sum / ;
             head->next = new ListNode(sum % );
             head = head->next;
             l2 = l2->next;
         }

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

         head->next = NULL;

         return dummy->next;
     }
 };

写得太长了，下面有短码的方法

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 addLists(ListNode l1, ListNode l2) {
         // write your code here
         ListNode dummy = new ListNode();
         ListNode tail = dummy;

         int carry = ;
         for (ListNode i = l1, j = l2; i != null || j != null; ) {
             int sum = carry;
             sum += (i != null) ? i.val : ;
             sum += (j != null) ? j.val : ;

             tail.next = new ListNode(sum % );
             tail = tail.next;

             carry = sum / ;
             i = (i == null) ? i : i.next;
             j = (j == null) ? j : j.next;
         }

         if (carry != ) {
             tail.next = new ListNode(carry);
         }
         return dummy.next;
     }
 }

参考了九章的代码

解法三：

 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
     ListNode preHead(), *p = &preHead;
     int extra = ;
     while (l1 || l2 || extra) {
         int sum = (l1 ? l1->val : ) + (l2 ? l2->val : ) + extra;
         extra = sum / ;
         p->next = new ListNode(sum % );
         p = p->next;
         l1 = l1 ? l1->next : l1;
         l2 = l2 ? l2->next : l2;
     }
     return preHead.next;
 }

参考了@ce2 的代码

解法四：

 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
         // 题意可以认为是实现高精度加法
         ListNode *head = new ListNode();
         ListNode *ptr = head;
         int carry = ;
         while (true) {
             if (l1 != NULL) {
                 carry += l1->val;
                 l1 = l1->next;
             }
             if (l2 != NULL) {
                 carry += l2->val;
                 l2 = l2->next;
             }
             ptr->val = carry % ;
             carry /= ;
             // 当两个表非空或者仍有进位时需要继续运算，否则退出循环
             if (l1 != NULL || l2 != NULL || carry != ) {
                 ptr = (ptr->next = new ListNode());
             } else break;
         }
         return head;
     }
 };

参考了九章的代码