【leetcode】2. 两数相加

 总体思路：

1.将两个链表里的数字相加：总左往右加，存入第三方链表L3里；

2.设置一个进位符 t，用来存储每位相加的进位信息；

3.对多出来单独的链表进行处理（只需考虑进位），接入到L3的后面。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *l3 = (struct ListNode*)malloc(sizeof(struct ListNode)),
                    *head = (struct ListNode*)malloc(sizeof(struct ListNode));
    int t = 0, m = 0;
    l3->val = 0;
    l3->next = NULL;
    head = l3;
    while (l1 != NULL && l2 != NULL) {
        struct ListNode* p = (struct ListNode*)malloc(sizeof(struct ListNode));
        p->val = ((l1->val + l2->val) + t) % 10;
        p->next = NULL;
        l3->next = p;
        l3 = p;
        t = (l1->val + l2->val + t) / 10;
        l1 = l1->next;
        l2 = l2->next;
    }
    if (l1 != NULL) {
        l3->next = l1;
        m = t;
        while ((l1->val + t) && l1->next != NULL) {
            m = t;
            t = (l1->val + t) / 10;
            l1->val = (l1->val + m) % 10;
            l1 = l1->next;
        }
        if ((l1->val + t) / 10 == 1) {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            p->val = 1;
            p->next = NULL;
            l1->val = (l1->val + t) % 10;
            l1->next = p;
        } else if (t == 0 && m == 1) {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            l1->val = l1->val + t;
        } else {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            l1->val = l1->val + t;
        }
    } else if (l2 != NULL) {
        l3->next = l2;
        m = t;
        while ((l2->val + t) && l2->next != NULL) {
            m = t;
            t = (l2->val + t) / 10;
            l2->val = (l2->val + m) % 10;
            l2 = l2->next;
        }
        if ((l2->val + t) / 10 == 1) {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            p->val = 1;
            p->next = NULL;
            l2->val = (l2->val + t) % 10;
            l2->next = p;
        } else {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            l2->val = l2->val + t;
        }
    } else {
        if (t == 1) {
            struct ListNode* p =
                (struct ListNode*)malloc(sizeof(struct ListNode));
            p->val = 1;
            p->next = NULL;
            l3->next = p;
        }
    }
    head = head->next;
    return head;
}