标签：des style blog color io ar strong sp div
Problem:
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
一开始就是想到先把第一个list转换成一个数，然后把第二个转换成第二个数，然后相加后，再把相加的值变成list。代码如下
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { int a = 0, b = 0, ans = 0, flag = 1; int index1 = 0, index2 = 0; ListNode *temp1, *temp2, *result; temp1 = l1; temp2 = l2; while(temp1->next != NULL ) { index1++; a += (temp1 -> next -> val) * (int)pow(10,index1); } while(temp2 -> next != NULL) { index2++; b += temp2 ->next -> val * (int)pow(10, index2); } ans = a + b; result -> val = ans%10; result -> next = NULL; flag = ans/10; while(flag) { ListNode *added = new ListNode(flag%10); result ->next = added; flag = flag/10; } return result; }
然后是Time Limit Exceed了。那就不能这样做，应该直接在链表相加。加到某个链表结束为止。要用中间变量记住当前的进位。如果最后进位不为零（也就是为1）的话，那还是需要记录的。代码贴出如下：
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode * ans = NULL, *last = NULL; int up = 0; while (l1 != NULL && l2 != NULL) { int tmp = l1->val + l2->val + up; up = tmp / 10; if (last == NULL) { ans = new ListNode(tmp % 10); last = ans; } else last = pushBack(last, tmp % 10); l1 = l1->next; l2 = l2->next; } while (l1 != NULL) { int tmp = l1->val + up; last = pushBack(last, tmp % 10); up = tmp / 10; l1 = l1->next; } while (l2 != NULL) { int tmp = l2->val + up; last = pushBack(last, tmp % 10); up = tmp / 10; l2 = l2->next; } if (up == 1) { ListNode * l = new ListNode(up); last->next = l; } return ans; } ListNode * pushBack(ListNode * last, int val) { ListNode * l = new ListNode(val); last->next = l; return l; } };
还是要感谢suool大神，改天一定要再做一次看看是不是真的掌握了。自己真的水平有限啊。不过只要肯努力，一天进步一点点就好。让cnblogs记录我的学习过程，come on！
标签：des style blog color io ar strong sp div
原文地址：http://www.cnblogs.com/higerzhang/p/4014737.html
