LeetCode 2. Add Two Numbers

题目

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

分析

变相大数加法, 较为不同的一点在于数据采用链表形式存储. 思路与普通大数加法相同, 只是结果储存于链表中. 计算过程中每计算一位, 分配一个 ListNode, 采用尾插法插入链表末尾. 值得注意的是, 最初

if (carry) {        struct ListNode *ptmp = malloc(sizeof(struct ListNode));        ptmp->val = carry;        p->next = ptmp;        p = ptmp;    }

后的代码为:

    prst = prst->next; // point to the first element    return prst;

提交后提示Runtime Error. 加入对链表头的内存释放后成功通过. 暂时不知道原因, 希望有高人可以帮忙解答.

    p = prst;    prst = prst->next; // point to the first element    free(p);    return prst;

解答

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode *pa = l1, *pb = l2, *prst, *p;    if (!pa || !pb)        return NULL;    p = prst = malloc(sizeof(struct ListNode));    prst->next = NULL; // head pointer, doesn't store data    int carry = 0;    while (l1 && l2) {        int a = l1->val;        int b = l2->val;        int tmp = a + b + carry;        carry = tmp / 10;        struct ListNode *ptmp = malloc(sizeof(struct ListNode));        ptmp->val = tmp % 10;        ptmp->next = p->next;        p->next = ptmp;        p = ptmp;        l1 = l1->next;        l2 = l2->next;    }    while (l1) {        int a = l1->val;        int tmp = a + carry;        carry = tmp / 10;        struct ListNode *ptmp = malloc(sizeof(struct ListNode));        ptmp->val = tmp % 10;        ptmp->next = p->next;        p->next = ptmp;        p = ptmp;        l1 = l1->next;    }    while (l2) {        int b = l2->val;        int tmp = b + carry;        carry = tmp / 10;        struct ListNode *ptmp = malloc(sizeof(struct ListNode));        ptmp->val = tmp % 10;        ptmp->next = p->next;        p->next = ptmp;        p = ptmp;        l2 = l2->next;    }    if (carry) {        struct ListNode *ptmp = malloc(sizeof(struct ListNode));        ptmp->val = carry;        p->next = ptmp;        p = ptmp;    }    p = prst;    prst = prst->next; // point to the first element    free(p);    return prst;}