crack the code interview 2.4

最新推荐文章于 2019-04-06 12:49:07 发布

yfdyyy

最新推荐文章于 2019-04-06 12:49:07 发布

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本文链接：https://blog.youkuaiyun.com/yfdyyy/article/details/17123297

本文介绍了一种利用链表实现大数加法的方法。两个链表分别存储两个数字，每个节点包含一个数字，且链表按逆序方式存储。通过遍历两个链表并逐位相加的方式实现加法运算，并将结果存储在一个新的链表中。

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You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

这是一个大数加法。有两个链表，每个链表的节点存放的是一个数字，其各位是链表的头节点，最后返回两个链表的和。

//You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

struct LinkNode
{
    LinkNode * next;
    int value;
}

LinkNode * BigAdd(LinkNode * n1, LinkNode * n2)
{
    LinkNode * sum = NULL;
    LinkNode * start = sum;
    
    LinkNode * h1 = n1;
    LinkNode * h2 = n2;
    int pre = 0;
    while (h1 != NULL && h2 != NULL)
    {
        int s = h1->value + h2->value + pre;
        LinkNode * n = new LinkNode(s, NULL);
        if (s >= 10)
        {
            n->value = (s) % 10;
            pre = s - (s/10) * 10;
        }
        if (sum == NULL)
        {
            sum = n;
        }
        else
            sum->next = n;
        h1 = h1->next;
        h2 = h2->next;
    }
    LinkNode * h = NULL;
    if (h1 != NULL)
        h = h1;
    else (h2 != NULL)
        h = h2;

    sum->next = h;
    while (pre != 0)
    {
        int s = pre + h->value;
        if (h != NULL)
        {
            if (s >= 10)
            {
                h->value = s % 10;
                pre = s - (s/10) * 10;
            }
        }
        else
        {
            LinkNode * n = new LinkNode(pre);
            pre = 0;
            h = n;
        }
        h = h->next;
    }
    return start;
}