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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
链表的对应节点加法。
/** * 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 *ans = new ListNode(0); ListNode *pos1 = l1; ListNode *pos2 = l2; int num = pos1->val + pos2->val, carry = 0; if(num > 9) { carry = 1; num -= 10; } ans->val = num; ans->next = nullptr; pos1 = pos1->next; pos2 = pos2->next; ListNode *before = ans; while(pos1 && pos2) { ListNode *tmp = new ListNode(0); auto num = carry + pos1->val + pos2->val; carry = 0; if(num > 9) { carry = 1; num -= 10; } tmp->val = num; before->next = tmp; before = tmp; pos1 = pos1->next; pos2 = pos2->next; } while(pos1) { ListNode *tmp = new ListNode(0); auto num = carry + pos1->val; carry = 0; if(num > 9) { carry = 1; num -= 10; } tmp->val = num; tmp->next = nullptr; before->next = tmp; before = tmp; pos1 = pos1->next; } while(pos2) { ListNode *tmp = new ListNode(0); auto num = carry + pos2->val; carry = 0; if(num > 9) { carry = 1; num -= 10; } tmp->val = num; tmp->next = nullptr; before->next = tmp; before = tmp; pos2 = pos2->next; } if(carry) { ListNode *tmp = new ListNode(carry); before->next = tmp; before = tmp; } return ans; } };
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原文地址:http://my.oschina.net/u/347565/blog/416435