【LeetCode算法练习（C语言）】Add Two Numbers

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题目：
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

链接：Add Two Numbers
解法：大数加法，时间O(n)

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode *p = (struct ListNode *)malloc(sizeof(struct ListNode));    p->val = 0;    p->next = NULL;    int add = 0;    struct ListNode *pp = p;    while (l1 && l2) {        if (l1->val + l2->val + add < 10) {            pp->val = l1->val + l2->val + add;            add = 0;        } else {            pp->val = l1->val + l2->val + add - 10;            add = 1;        }        struct ListNode *t = (struct ListNode *)malloc(sizeof(struct ListNode));        t->val = 0;        t->next = NULL;        pp->next = t;        pp = t;        l1 = l1->next;        l2 = l2->next;    }    while (l1) {        if (l1->val + add < 10) {            pp->val = l1->val + add;            add = 0;        } else {            pp->val = l1->val + add - 10;            add = 1;        }        struct ListNode *t = (struct ListNode *)malloc(sizeof(struct ListNode));        t->val = 0;        t->next = NULL;        pp->next = t;        pp = t;        l1 = l1->next;    }    while (l2) {        if (l2->val + add < 10) {            pp->val = l2->val + add;            add = 0;        } else {            pp->val = l2->val + add - 10;            add = 1;        }        struct ListNode *t = (struct ListNode *)malloc(sizeof(struct ListNode));        t->val = 0;        t->next = NULL;        pp->next = t;        pp = t;        l2 = l2->next;    }    if (add) {        pp->val = 1;    } else {        struct ListNode *tmp = p;        while (tmp->next != pp) {            tmp = tmp->next;        }        tmp->next = NULL;        free(pp);    }    return p;}

Runtime: 45 ms

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