题目：
示例：
思路：
有点相似于：链表相加II，这道题我们仍然有进位，但不同的是，链表相加我们选择了开辟新节点，这道题我们选择反转两次链表，开始一次，结束一次。
单链表的翻转详细讲解：反转一个单链表（<---点击可看详解）
题解代码：
struct ListNode* reverse(struct ListNode* head) { struct ListNode* n1 = NULL; struct ListNode* n2 = head; struct ListNode* n3 = NULL; if(n2) n3 = n2->next; while (n2) { n2->next = n1; n1 = n2; n2 = n3; if (n3) n3 = n3->next; } return n1; } struct ListNode* doubleIt(struct ListNode* head) { struct ListNode* prev = NULL; struct ListNode* Back = reverse(head); head = Back; int add = 0; while(Back) { int temp = Back->val * 2 + add; add = temp / 10; prev = Back; Back->val = temp % 10; Back = Back->next; } if(add > 0) { struct ListNode* newnode = malloc(sizeof(struct ListNode)); newnode->val = add; newnode->next = NULL; prev->next = newnode; } head = reverse(head); return head; }
