题解 | #链表相加(二)#

链表相加(二)

https://www.nowcoder.com/practice/c56f6c70fb3f4849bc56e33ff2a50b6b

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
 *
 * 
 * @param head1 ListNode类 
 * @param head2 ListNode类 
 * @return ListNode类
 */

typedef struct ListNode LN;

struct ListNode* addInList(struct ListNode* head1, struct ListNode* head2 ) 
{
    int flag = 0;
    LN* newhead = NULL;
    int* arr1 = (int*)malloc(sizeof(int)*1000000);
    int* arr2 = (int*)malloc(sizeof(int)*1000000);
    LN* pmove1 = head1;
    int num1 = 0;
    LN* pmove2 = head2;
    int num2 = 0;
    while(pmove1)
    {
        arr1[num1++] = pmove1->val;
        pmove1 = pmove1->next;
    }
    while(pmove2)
    {
        arr2[num2++] = pmove2->val;
        pmove2 = pmove2->next;
    }
    num1--;
    num2--;
    while(num1>=0||num2>=0)
    {
        LN* newnode = (LN*)malloc(sizeof(LN));
        int x = num1>=0?arr1[num1]:0;
        int y = num2>=0?arr2[num2]:0;
        int add = (x+y+flag)%10;
        newnode->val = add;
        newnode->next = newhead;
        newhead = newnode;
        if(x+y+flag>=10)
        {
            flag = 1;
        }
        else 
        {
            flag = 0;
        }
        if(num1>=0)
        {
            num1--;
        }
        if(num2>=0)
        {
            num2--;
        }
    }
    if(flag)
    {
        LN* newnode = (LN*)malloc(sizeof(LN));
        newnode->next = NULL;
        newnode->val = 1;
        newnode->next = newhead;
        newhead = newnode;
    }
    return newhead;
}

全部评论

相关推荐

评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客企业服务