标签：single output 部分 node pre tput null self code
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
想法：这题思路很清晰，逐个节点相加即可，但是需要注意到结果大于9，会存在进位的问题，因而需要设置变量jinwei来保存这个进位信息。还得考虑到两个链表非等长的情况，这时处理完两个链表共有的部分之后，单独处理剩下的链表，这个时候仍有会有进位的信息存在。
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode* p = l1; struct ListNode* q = l2; struct ListNode* temp = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* r = temp; int jinwei = 0;//保存进位信息
//考虑两个链表共有的部分 while(p && q){ struct ListNode* t = (struct ListNode*)malloc(sizeof(struct ListNode)); t->val = (p->val + q->val + jinwei) % 10; jinwei = (p->val + q->val + jinwei) / 10; temp->next = t; temp = t; p = p->next; q = q->next; } while(p){ struct ListNode* t = (struct ListNode*)malloc(sizeof(struct ListNode)); t->val = (p->val + jinwei)%10; jinwei = (p->val + jinwei)/10; temp->next = t; temp = t; p = p->next; } while(q){ struct ListNode* t = (struct ListNode*)malloc(sizeof(struct ListNode)); t->val = (q->val+jinwei)%10; jinwei = (q->val+jinwei)/10; temp->next = t; temp = t; q = q->next; }
//如若所有节点处理完之后，仍然还有进位信息存在，则需要另外申请节点储存该信息 if(jinwei!=0){ struct ListNode* t = (struct ListNode*)malloc(sizeof(struct ListNode)); t->val = jinwei; temp->next =t; temp = t; } temp->next = NULL; return r->next; }
标签：single output 部分 node pre tput null self code
原文地址：https://www.cnblogs.com/tingweichen/p/9861268.html
