题目: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:
从头到后的顺序,注意有进位,一个小技巧就是先是在程序判断两边都还有数字,均不为空,再判断谁已经“ 出局 ”。
代码:
/** * 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 *dummy=new ListNode(0); ListNode *tail=dummy; int flag=0; while(l1||l2){ int m=0; if(l1){ m+=l1->val; l1=l1->next; } if(l2){ m+=l2->val; l2=l2->next; } m=m+flag; ListNode *tmp=new ListNode(m%10); flag=m/10; tail->next=tmp; tail=tail->next; } if(flag){ ListNode *tmp=new ListNode(flag); tail->next=tmp; } return dummy->next; } };