Add Two Numbers

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#C++ #编程 #leetcode

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本文介绍了如何使用C++解决两个非负整数以链表形式表示并逆序存储的相加问题。通过定义链表节点结构，并实现将两个链表相加的功能，最终返回一个表示相加结果的链表。代码示例清晰地展示了从链表头节点开始遍历，进行数值加法和进位处理的过程。

题目：

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

解法：

题目很简单，不多说了。代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(l1==NULL)
            return l2;
        if(l2==NULL)
            return l1;
        int sum,bit;
        sum = (l1->val+l2->val)%10;
        bit = (l1->val+l2->val)/10;
        ListNode * head=new ListNode(sum);
        ListNode * tmp = head;
        while(l1->next && l2->next)
        {
            sum = (l1->next->val+l2->next->val+bit)%10;
            bit = (l1->next->val+l2->next->val+bit)/10;
            tmp->next = new ListNode(sum);
            tmp = tmp->next;     
            l1 = l1->next;
            l2 = l2->next;       
        }
        while(l1->next)
        {
            sum = (l1->next->val+bit)%10;
            bit = (l1->next->val+bit)/10;
            tmp->next = new ListNode(sum);
            tmp = tmp->next; 
            l1 = l1->next;
        }
        while(l2->next)
        {
            sum = (l2->next->val+bit)%10;
            bit = (l2->next->val+bit)/10;
            tmp->next = new ListNode(sum);
            tmp = tmp->next; 
            l2 = l2->next;
        }
        if(bit>0)
        {
            tmp->next = new ListNode(bit);
        }
        return head;
    }
};