Lintcode Add Two Numbers II

最新推荐文章于 2018-02-04 22:44:00 发布

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#lintcode #AddTwoNumb

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本文介绍了一种链表加法算法的实现方法，通过反转链表、逐位相加并处理进位，最后再次反转得到结果。适用于需要进行链表数值运算的场景。

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Example

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

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 class Solution {
 public:
     /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
     void ListReverse(ListNode*& list)  //List Reverse
     {
         if (list == NULL || list->next == NULL)  //特殊情况
             return;
         ListNode* prev = list, *p = list->next, *pnext = p->next; // 记录三个节点的位置
         prev->next = NULL;

         while (p != NULL)
         {
             pnext = p->next;  //记录下一位置
             p->next = prev;  //指针前指
             prev = p; 
             p = pnext; // 下一位置继续循环
         }
         list = prev; // 首节点为原来的尾节点

     }

     ListNode* Add(ListNode* l1, ListNode* l2)
     {
         ListNode* sum = NULL, *p1 = NULL, *pf = sum;

         // 先把所有的对应位置都进行相加 保存到sum 中

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

             if (sum == NULL)
             {
                 sum = new ListNode(l1->val + l2->val);
                 pf = sum;
             }
             else
             {
                 pf->next = new ListNode(l1->val + l2->val);
                 pf = pf->next;
             }
             l1 = l1->next;
             l2 = l2->next;
         }
         while (l1 != NULL)
         {
             if (sum == NULL) //sum is Null, sum = new listnode
             {
                 sum = new ListNode(l1->val);
                 pf = sum;
             }
             else
             {
                 pf->next = new ListNode(l1->val);
                 pf = pf->next;
             }
             l1 = l1->next;
         }
         while (l2 != NULL)
         {
             if (sum == NULL)
             {
                 sum = new ListNode(l2->val);
                 pf = sum;
             }
             else
             {
                 pf->next = new ListNode(l2->val);
                 pf = pf->next;
             }
             l2 = l2->next;
         }
         p1 = sum;
         bool flag = false;  //进位记录
         while (p1 != NULL)
         {
             if (flag == true)
                 p1->val++;

             if (p1->val >= 10)
             {
                 flag = true;
                 p1->val -= 10;
             }
             else
                 flag = false;
             if (flag == true && p1->next == NULL)
             {
                 p1->next = new ListNode(1);
                 flag = false;
             }
             p1 = p1->next;
         }
         return sum;
     }
     ListNode *addLists2(ListNode *l1, ListNode *l2) {
         // write your code here
         ListReverse(l1);
         ListReverse(l2);
         ListNode* out = Add(l1, l2);
         ListReverse(out);
         return out;
     }
 };