leetcode 2. Add Two Numbers
https://leetcode.com/problems/add-two-numbers/
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：方法比较简单，直接将每个节点的值相加取10的余数，然后除10为进位。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { 12 ListNode *t1 = l1; 13 ListNode *t2 = l2; 14 int k=0; 15 int a1,a2,t=0; 16 ListNode *pre =NULL; 17 ListNode *result = NULL; 18 while(t1&&t2) { 19 a1=t1->val;t1=t1->next; 20 a2=t2->val;t2=t2->next; 21 ListNode *temp = new ListNode((a1+a2+t)%10); 22 t = (a1+a2+t)/10; 23 if(pre) pre->next = temp; 24 else result = temp; 25 pre=temp; 26 } 27 while(t1){ 28 int a = t1->val;t1=t1->next; 29 ListNode *temp = new ListNode((a+t)%10); 30 t = (a+t)/10; 31 if(pre) pre->next = temp; 32 else result = temp; 33 pre = temp; 34 } 35 while(t2){ 36 int a = t2->val;t2=t2->next; 37 ListNode *temp = new ListNode((a+t)%10); 38 t = (a+t)/10; 39 if(pre) pre->next = temp; 40 else result = temp; 41 pre = temp; 42 } 43 if(t) { 44 ListNode *temp = new ListNode(t); 45 pre->next = temp; 46 } 47 return result; 48 } 49 50 };
在leetcode上时间为36ms。
