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
AC代码如下：
ListNode* add(ListNode* l1, ListNode* l2,int len1,int len2){ if(len1 <len2) return add(l2,l1,len2,len1); //len1>=len2; ListNode* tmp1=l1; ListNode* tmp2=l2; int jin=0; while(tmp2!=NULL) { int now = (tmp1->val+tmp2->val)+jin; if(now>=10) { jin = now/10; now = now-10; } else { jin= 0; } tmp1->val = now; tmp1=tmp1->next; tmp2=tmp2->next; } for(;jin>0,tmp1!=NULL;tmp1=tmp1->next) { int now = jin + tmp1->val; if(now >= 10) { jin = now/10; now = now-10; } else { jin=0; } tmp1->val = now; } if(tmp1==NULL&&jin>0) { //ListNode node(jin); ListNode* nodenew = new ListNode(jin); ListNode *tmp=l1; for(;tmp->next!=NULL;tmp=tmp->next) ; tmp->next = nodenew; } return l1; } ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if(l1 == NULL && l2 == NULL) return NULL; else if(l1 == NULL) return l2; else if(l2 == NULL) return l1; else { int len1=0,len2=0; ListNode* tmp; for(tmp=l1;tmp!=NULL;tmp=tmp->next) { len1++; } for(tmp=l2;tmp!=NULL;tmp=tmp->next) { len2++; } return add(l1,l2,len1,len2); } }
别人写的，也差不多。主要需要考虑进位的事情。有多种情况。两个链表长度相等时，只要最后有进位，就直接放到后面。如果不等，则把长度小的数加到长度较大的数上，然后判断进位。
原文：http://www.cnblogs.com/ww-jin/p/4445614.html