LeetCode--Add Two Numbers

问题

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

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* create() {    struct ListNode *node = (struct ListNode *) malloc(sizeof(struct ListNode));    if (node == NULL) {        printf("allocation failed\n");        exit(0);    }    node->next = NULL;    return node;}int add(int va1, int va2, int *val) {    int flag = 0;    *val = va1 + va2;    if (*val >= 10) {        flag = 1;        *val = *val % 10;    }    return flag;}struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {    struct ListNode *head = create();    int flag = 0;    int val = 0;    flag = add(l1->val, l2->val, &val);    head->val = val;    struct ListNode *p = head;    l1 = l1->next;    l2 = l2->next;    while(l1 != NULL && l2 != NULL) {        flag = add(l1->val + flag, l2->val, &val);        struct ListNode *node = create();        node->val = val;        p->next = node;        p = node;        l1 = l1->next;        l2 = l2->next;    }    while(l1 != NULL) {        struct ListNode *node = create();        if (flag == 1) {            flag = add(flag, l1->val, &val);        } else {            val = l1->val;        }        node->val = val;        p->next = node;        p = node;        l1 = l1->next;    }    while(l2 != NULL) {        struct ListNode *node = create();        if (flag == 1) {            flag = add(flag, l2->val, &val);        } else {            val = l2->val;        }        node->val = val;        p->next = node;        p = node;        l2 = l2->next;    }    if (flag == 1) {        struct ListNode *node = create();        node->val = 1;        p->next = node;    }    return head;}

性能