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

 

这道并不是什么难题，算法很简单，链表的数据类型也不难。就是建立一个新链表，然后把输入的两个链表从头往后撸，每两个相加，添加一个新节点到新链表后面，就是要处理下进位问题。还有就是最高位的进位问题要最后特殊处理一下。代码如下：

 

C++ 解法： 

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        ListNode *res = new ListNode(-1);        ListNode *cur = res;        int carry = 0;        while (l1 || l2) {            int n1 = l1 ? l1->val : 0;            int n2 = l2 ? l2->val : 0;            int sum = n1 + n2 + carry;            carry = sum / 10;            cur->next = new ListNode(sum % 10);            cur = cur->next;            if (l1) l1 = l1->next;            if (l2) l2 = l2->next;        }        if (carry) cur->next = new ListNode(1);        return res->next;    }};

 

Java 解法：

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode dummy = new ListNode(-1);        ListNode cur = dummy;        int carry = 0;        while (l1 != null || l2 != null) {            int d1 = l1 == null ? 0 : l1.val;            int d2 = l2 == null ? 0 : l2.val;            int sum = d1 + d2 + carry;            carry = sum >= 10 ? 1 : 0;            cur.next = new ListNode(sum % 10);            cur = cur.next;            if (l1 != null) l1 = l1.next;            if (l2 != null) l2 = l2.next;        }        if (carry == 1) cur.next = new ListNode(1);        return dummy.next;    }}

 

在CareerCup上的这道题还有个Follow Up，把链表存的数字方向变了，原来是表头存最低位，现在是表头存最高位，请参见我的另一篇博客 。

 

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