两个链表的和

容易  两个链表的和

21 %

通过

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

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Cracking The Coding Interview 链表 高精度

---

---

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2
     */
    /**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/

int add(int a, int b, int &carry)
{
    int c = a + b + carry;
   
    carry = c / 10;
    return c % 10;
}

ListNode *addLists(ListNode *l1, ListNode *l2)
{
    ListNode *head = NULL, *rear = NULL;
   
    if (l1 == NULL) return l2;
    if (l2 == NULL) return l1;
   
    int carry = 0;
   
    while (l1 != NULL || l2 != NULL)
    {
        int val = 0;
       
        if (l1 == NULL)
        {
            val = add(0, l2->val, carry);
            l2 = l2->next;
        }
        else if (l2 == NULL)
        {
            val = add(l1->val, 0, carry);
            l1 = l1->next;
        }
        else
        {
            val = add(l1->val, l2->val, carry);
            l1 = l1->next;
            l2 = l2->next;
        }
       
        ListNode *new_node = new ListNode(val);
       
        if (head == NULL) head = new_node;
        if (rear == NULL)
        {
            rear = new_node;
        }
        else
        {
            rear->next = new_node;
            rear = rear->next;
        }
    }
   
    if (carry != 0 && rear != NULL)
    {
        rear->next = new ListNode(carry);
    }
   
    return head;
}
};