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题目：

示例： 

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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） 

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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;
}

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