LeetCode Add Two Numbers两个链表数值相加

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

相当于无限加了，就是用了链表来存储数据。

本题就是考对链表的操作和模拟加法。

第一种方法：

这个跟合并两个有序链表成一个有序链表的程序结构是一样的。

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {  
        ListNode *sumList = new ListNode(-1);  
        ListNode *cur = sumList;  
  
        int carry = 0;  
        while (l1 != nullptr || l2 != nullptr || carry != 0)   
        {  
            int l1val = l1 == nullptr   0 : l1->val;  
            int l2val = l2 == nullptr   0 : l2->val;  
            int sum = l1val + l2val + carry;  
            carry = sum / 10;  
            sum %= 10;  
  
            cur->next = new ListNode(sum);  
            cur = cur->next;  
  
            l1 = l1 == nullptr   nullptr : l1->





next;  
            l2 = l2 == nullptr   nullptr : l2->next;  
        }  
  
        ListNode *t = sumList->next;  
        delete sumList;  
        return t;  
    }  

x

这样也许简洁一点，逻辑是差不多的，就是处理数据有点不一样。

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {  
        ListNode *sumList = new ListNode(-1);  
        ListNode *cur = sumList;  
  
        int carry = 0;  
        while (l1 != nullptr || l2 != nullptr || carry != 0)   
        {  
            int l1val = l1 == nullptr   0 : l1->val;  
            int l2val = l2 == nullptr   0 : l2->val;  
            int sum = l1val + l2val + carry;  
            carry = sum / 10;  
            sum %= 10;  
  
            cur->next = new ListNode(sum);  
            cur = cur->next;  
  
            l1 = l1 == nullptr   nullptr : l1->next;  
            l2 = l2 == nullptr   nullptr : l2->next;  
        }  
  
        ListNode *t = sumList->next;  
        delete sumList;  
        return t;  
    }