167. Add Two Numbers【easy】

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse 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 7->1->6 + 5->9->2. That is, 617 + 295.

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

Given 3->1->5 and 5->9->2, return 8->0->8.

题意

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

解法一：

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode addLists(ListNode l1, ListNode l2) {
         if(l1 == null && l2 == null) {
             return null;
         }

         ListNode head = new ListNode(0);
         ListNode point = head;
         int carry = 0;
         while(l1 != null && l2!=null){
             int sum = carry + l1.val + l2.val;
             point.next = new ListNode(sum % 10);
             carry = sum / 10;
             l1 = l1.next;
             l2 = l2.next;
             point = point.next;
         }

         while(l1 != null) {
             int sum =  carry + l1.val;
             point.next = new ListNode(sum % 10);
             carry = sum /10;
             l1 = l1.next;
             point = point.next;
         }

         while(l2 != null) {
             int sum =  carry + l2.val;
             point.next = new ListNode(sum % 10);
             carry = sum /10;
             l2 = l2.next;
             point = point.next;
         }

         if (carry != 0) {
             point.next = new ListNode(carry);
         }
         return head.next;
     }
 }

中规中矩的解法

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) {
         ListNode dummy = new ListNode(0);
         ListNode tail = dummy;

         int carry = 0;
         for (ListNode i = l1, j = l2; i != null || j != null; ) {
             int sum = carry;
             sum += (i != null) ? i.val : 0;
             sum += (j != null) ? j.val : 0;

             tail.next = new ListNode(sum % 10);
             tail = tail.next;

             carry = sum / 10;
             i = (i == null) ? i : i.next;
             j = (j == null) ? j : j.next;
         }

         if (carry != 0) {
             tail.next = new ListNode(carry);
         }
         return dummy.next;
     }
 }

比较简明的写法，且使用了dummy节点，参考@NineChapter 的代码

解法三：

 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) {
         ListNode root = new ListNode(-1);
         ListNode result = root;
         int carry = 0;        

         while( l1 != null || l2 != null || carry == 1){
             int value = 0;
             if(l1 != null){
                 value += l1.val;
                 l1 = l1.next;
             }
             if( l2 != null){
                 value += l2.val;
                 l2 = l2.next;
             }

             value += carry;
             root.next = new ListNode(value % 10);
             carry = value / 10;

             root = root.next;

         }

         return result.next;
     }
 }

解法四：

 class Solution {
 public:
     ListNode* addLists(ListNode* l1, ListNode* l2) {
         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;
     }
 };

非常简明的代码，参考@NineChapter 的代码