Leetcode : Add Two Numbers

最新推荐文章于 2019-02-21 19:00:24 发布

原创最新推荐文章于 2019-02-21 19:00:24 发布 · 565 阅读

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本文介绍了一种使用链表实现加法的方法，输入为两个非负数的链表表示，链表中的数字按逆序存储。文章详细解释了如何处理进位问题，并提供了完整的代码示例。

Add Two NumbersNov 1 '11

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

Note: 不知道大家是不是都看懂题目了，我就看了好久！就例子来说，其实就是342 + 465 = 807，但是是用链表倒序存储的。看懂题目之后，做起来就比较简单了，唯一需要注意的地方就是“进位”的问题！

代码如下：

ListNode *addTwoNumbers(ListNode *first, ListNode *second)
{   
    int carry = 0; // 进位
    
    ListNode* pHead = NULL;
    ListNode* pTail = NULL;
    
    while (first && second)
    {
        int x = first->val + second->val + carry; 
        
        ListNode* pNext = new ListNode(x%10);
        carry = x/10;
        if (!pHead)
        {
            pHead = pNext;
            pTail = pNext;
        }
        else
        {
            pTail->next = pNext;
            pTail = pNext;
        }
        
        first = first->next;
        second = second->next;
    }
    

    while (second)
    {
        int x = second->val + carry;
        carry = x/10;
        
        ListNode* pNext = new ListNode(x%10);
        
        if (!pHead)
        {
            pHead = pNext;
            pTail = pNext;
        }
        else
        {
            pTail->next = pNext;
            pTail = pNext;
        }
        
        second = second->next;
    }
    
    while (first)
    {
        int x = first->val + carry;
        carry = x/10;
        
        ListNode* pNext = new ListNode(x%10);
        
        if (!pHead)
        {
            pHead = pNext;
            pTail = pNext;
        }
        else
        {
            pTail->next = pNext;
            pTail = pNext;
        }
        
        first = first->next;
    }
    
    if (carry)
    {
        ListNode* pNext = new ListNode(carry);
        pTail->next = pNext;
        pTail = pNext;
    }
        
    return pHead;
}