标签：des style blog class code c
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
思路：将两个链表从表头开始逐个相加，遇到大于10的则向后一节点进1。使用两个指针分别指向两个链表，且进位值用mod记录，如果一个链表没有结束，则将这个链表的剩余部分添加进去，注意进位。如果都结束了，但最后一位出现了进位，则要添加一个结点到新链表中。
/** * 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) { if(l1==NULL) return l2; else if(l2==NULL) return l1; ListNode *lHead=NULL; ListNode *pList=NULL; int mod=0; while(l1!=NULL&&l2!=NULL) { int temp=l1->val+l2->val+mod; mod=temp/10; temp%=10; if(lHead==NULL) { lHead=new ListNode(temp); pList=lHead; } else { pList->next=new ListNode(temp); pList=pList->next; } l1=l1->next; l2=l2->next; } while(l1!=NULL) { int temp=l1->val+mod; mod=temp/10; temp%=10; if(lHead==NULL) { lHead=new ListNode(temp); pList=lHead; } else { pList->next=new ListNode(temp); pList=pList->next; } l1=l1->next; } while(l2!=NULL) { int temp=l2->val+mod; mod=temp/10; temp%=10; if(lHead==NULL) { lHead=new ListNode(temp); pList=lHead; } else { pList->next=new ListNode(temp); pList=pList->next; } l2=l2->next; } if(mod!=0)//注意这里。 pList->next=new ListNode(mod); return lHead; } };
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标签：des style blog class code c
原文地址：http://www.cnblogs.com/awy-blog/p/3725062.html
