LeetCode-002.Add Two Numbers

题目描述：

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路：

链表每一个结点为整数的一位数字，且链表为整数的倒序排列，即从链表头开始是整数最低位数字。因此每位数字相加即可，可能会出现三种情况：

1. l1、l2当前结点均存在，则结果为两结点值和前一位进位值相加对10取余，且要计算对后面的进位。
2. l1、l2当前结点只有一个存在，则结果为该结点值和前一位进位值相加对10取余，且要计算对后面的进位。
3. l1、l2当前节点均不存在，则判断前一位进位值。若为1，则结果为1；若为0则计算结束。

C语言解法：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    int carry = 0;  
    struct ListNode *head, *sum;
    head = (struct ListNode *)malloc(sizeof(struct ListNode));
    sum = head;
    sum->next = NULL;
    while(l1 != NULL || l2 != NULL || carry){
        if(l1 != NULL && l2 != NULL){
            sum->val = (l1->val + l2->val + carry) % 10;
            carry = (l1->val + l2->val + carry) / 10;
        } else if(l1 == NULL && l2 != NULL){
            sum->val = (l2->val + carry) % 10;
            carry = (l2->val + carry) / 10;
        } else if(l2 == NULL && l1 !=NULL){
            sum->val = (l1->val + carry) % 10;
            carry = (l1->val + carry) / 10;
        } else if(carry){
            sum->val = carry;
            carry = 0;
        }

        if(l1 != NULL) l1 = l1->next;
        if(l2 != NULL) l2 = l2->next;

        if(l1!=NULL || l2!=NULL || carry){
            struct ListNode* t;
            t = (struct ListNode *)malloc(sizeof(struct ListNode));
            t->next = sum->next;
            sum->next = t;
            sum = sum->next;    
        }
    }
    return head;
}

C++解法：

/**
 * 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 carry = 0;
        ListNode *head, *sum;
        head = new ListNode(0);
        sum = head;
        while(l1 != NULL || l2 != NULL || carry){
            if(l1 !=NULL && l2 != NULL){
                *sum = ListNode((l1->val + l2->val + carry) % 10);
                carry = (l1->val + l2->val + carry) / 10;
            } else if(l1 != NULL && l2 == NULL){
                *sum = ListNode((l1->val + carry) % 10);
                carry = (l1->val + carry) / 10;
            } else if(l1 == NULL && l2 != NULL){
                *sum = ListNode((l2->val + carry) % 10);
                carry = (l2->val + carry) / 10;
            } else if(carry){
                *sum = ListNode(carry);
                carry = 0;
            }

            if(l1 != NULL) l1 = l1->next;
            if(l2 != NULL) l2 = l2->next;

            if(l1 != NULL || l2 != NULL || carry){
                sum->next = new ListNode(0);
                sum = sum->next;
            }
        }
        return head;
    }
};