LeetCode OJ：Add Two Numbers (相加链表之数)

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

将两个链表上的数相加就可以，大于10进一位，注意下相加时候的细节就可以了，我这里

吧prev节点记录下来，这样最后多生成节点的时候便于将最后一个多余的节点删除：

 /**

  * Definition for singly-linked list.

  * struct ListNode {

  *     int val;

  *     ListNode *next;

  *     ListNode(int x) : val(x), next(NULL) {}

  * };

  */

 class Solution {

 public:

     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

         ListNode * curr = new ListNode();

         ListNode * root = curr;

         ListNode * prev = curr;

         int currVal = ;

         while(l1 != NULL && l2 != NULL){

             currVal = l1->val + l2->val;

             curr->val += currVal;

             curr->next = new ListNode(curr->val/);

             curr->val%=;

             prev = curr;

             curr = curr->next;

             l1 = l1->next;

             l2 = l2->next;

         }

         while(l1 != NULL){

             curr->val += l1->val;

             curr->next = new ListNode(curr->val/);

             curr->val %= ;

             prev = curr;

             curr = curr->next;

             l1 = l1->next;

         }

         while(l2 != NULL){

             curr->val += l2->val;

             curr->next = new ListNode(curr->val/);

             curr->val %= ;

             prev = curr;

             curr = curr->next;

             l2 = l2->next;

         }

         if(curr->val == ){

             prev->next = NULL;

             delete curr;

         }

         return root;

     }

 };

感觉写的有点麻烦，应该有很多的重复代码可以改正，但是我暂时找不出来了，先这样吧。

更新下，以前脑子抽了写出了那样的代码。其实三个while循环都可以放到一个while中，下面用java在写一起，方法还是类似的：

 /**

  * Definition for singly-linked list.

  * public class ListNode {

  *     int val;

  *     ListNode next;

  *     ListNode(int x) { val = x; }

  * }

  */

 public class Solution {

     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

         ListNode tmp = new ListNode(0);

         ListNode head = tmp;

         int carry = 0;

         while(l1!=null || l2!=null || carry != 0){

             int val = ((l1 != null)?l1.val:0) + ((l2!=null)?l2.val:0) + carry;

             carry = carry/10 + val/10;

             val %= 10;

             tmp.next = new ListNode(val);

             tmp = tmp.next;

             l1 = l1!=null ? l1.next : l1;

             l2 = l2!=null ? l2.next : l2;

         }

         return head.next;

     }

 }

巴特西

LeetCode OJ：Add Two Numbers (相加链表之数)

最新文章

热门文章