两数相加 2 、445

 

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    if(!l1 || !l2) {
        return NULL;
    }
    struct ListNode * ret = (struct ListNode *)calloc(1, sizeof(struct ListNode));
    int lastnum = 0;
    struct ListNode * result = ret;
    while (l1 != NULL || l2 != NULL) {
        int val1 = l1 != NULL ? l1->val : 0;
        int val2 = l2 != NULL ? l2->val : 0;
        ret -> val = (val1 + val2 + lastnum) % 10;
        printf("val:%d \n", ret -> val);
        lastnum = (val1 + val2 + lastnum) / 10;
        if (l1) l1 = l1->next;
        if (l2) l2 = l2->next;
        if (l1 != NULL || l2 != NULL) {
            struct ListNode * next = (struct ListNode *)calloc(1, sizeof(struct ListNode));
            ret->next = next;
            ret = ret->next;
        }

    }
    if (lastnum) {
         struct ListNode * next = (struct ListNode *)calloc(1, sizeof(struct ListNode));
         next->val = lastnum;
         printf("val:%d \n", next->val);
         ret->next = next;
    }
    return result;
}

 

 

 

 

int getlength(struct ListNode* node) 
{
    if (!node) {
        return 0;
    }
    int length = 0;
    while (node) {
        length++;
        node= node->next;
    }
    return length;
}
static struct ListNode* extendList(struct ListNode *node, int num)
{
    struct ListNode* ret = (struct ListNode* )calloc(1, sizeof(struct ListNode));
    struct ListNode* result = ret;
    for (int i = 0; i < num - 1; i++) {
        ret->next = (struct ListNode* )calloc(1, sizeof(struct ListNode));
        ret = ret->next;
    }
    ret->next = node;
    return result;
}
int calculate(struct ListNode** result, struct ListNode* num1, struct ListNode* num2)
{
    
    if (!num1 || !num2) {
        return 0;
    } else {
        int num = calculate(result, num1->next, num2->next);
        int sum = num1->val + num2->val + num;
        struct ListNode *tmp = (struct ListNode* )calloc(1, sizeof(struct ListNode));
        tmp->val = sum % 10;
        //printf("val:%d\n", tmp->val);
        tmp->next = *result;
        *result = tmp;
        return sum / 10;
    }
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    if (!l1 && !l2) {
        return NULL;
    }
    int len1 = getlength(l1);
    int len2 = getlength(l2);
    if (len1 > len2) {
        l2 = extendList(l2, len1 - len2);
    } else if( len2 > len1) {
        l1 = extendList(l1, len2 - len1);
    }
    int len = len1 > len2 ? len1 : len2;
    struct ListNode* ret = NULL;
    int value = calculate(&ret, l1, l2);

    if(value) {
        struct ListNode *tmp = (struct ListNode* )calloc(1, sizeof(struct ListNode));
        tmp->val = value;
        tmp->next = ret;
        ret = tmp;
    }
    return ret;

}

leetcode 445 lension learn：

1、 打印debug 时会导致链表头指针偏离， 打印debug正确后需要删除该部分，才能获取正确结果。 代码编写拷贝时出现入参 或者错别字的地基错误， 需要格外注意负责浪费过多的时间。

2、相加运算时需要记录进位信息， 且最高位的进位信息需要单独处理

3、 迭代运用的不够熟练， 特别是迭代函数中链表新元素创建之后需要赋值的过程。含义为创建新的节点后，原来的链表首节点作为新建节点的next， 再把新建节点作为返回链表result 的首节点。

        struct ListNode *tmp = (struct ListNode* )calloc(1, sizeof(struct ListNode));
        tmp->val = sum % 10;
        //printf("val:%d\n", tmp->val);
        tmp->next = *result;
        *result = tmp;
        return sum / 10;