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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
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; /* 最直白的想法就是将两个字符串翻转, 然后相加,得到的结果再发转 现在的解法,直接从左到右相加,然后向右进位 */ class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode* ptr = new ListNode(0); ListNode* head = ptr; int tmp_carry = 0; int sum; while( l1 || l2 ) { sum = 0; if( l1 ) { sum += l1->val; l1 = l1->next; } if( l2 ) { sum += l2->val; l2 = l2->next; } ptr->next = new ListNode( (sum + tmp_carry)%10 ); ptr = ptr->next; tmp_carry = (sum + tmp_carry)/10; } if( tmp_carry ) { ptr->next = new ListNode( tmp_carry ); } return head->next; } };
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原文地址:http://www.cnblogs.com/aceg/p/4423425.html