【LeetCode】2. Add Two Numbers

问题描述：

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

用一个链表表示一个非负数，链表尾的数为最高位，输入两个链表，求相加后的数对应的链表。

解题思路：

这道题很简单，需要注意的是输入的两个链表长度可能不同，

逐位相加。

注意结果的链表头和链表尾（最高位可能有进位）的处理就可以了。

参考代码：

非递归版本：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        int s = l1->val + l2->val;        int carry = s / 10;        ListNode *head = new ListNode(s % 10);        ListNode *r = head;        l1 = l1->next;        l2 = l2->next;        while(l1 || l2 || carry){            int w = carry;            if (l1){                w += l1->val;                l1 = l1->next;            }            if (l2){                w += l2->val;                l2 = l2->next;            }                        r->next = new ListNode(w % 10);            r = r->next;            carry = w / 10;                    }        return head;    }};

递归版本：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        return add(l1,l2,0);    }    ListNode* add(ListNode* l1, ListNode* l2, int carry){        if (!l1 && !l2 && !carry)return 0;        int s = 0;        if (l1)s += l1 -> val;        if (l2)s += l2 -> val;        s += carry;        ListNode *node = new ListNode(s % 10);        if (l1)l1 = l1->next;        if (l2)l2 = l2->next;        node->next = add(l1,l2,s / 10);        return node;    }};