链表 -- 2.2.1 Add Two Numbers -- 图解

描述
You are given two linked lists representing two non-negative numbers. e 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

class Solution{
public:ListNode *addTwoNumbers(ListNode *l1,ListNode *l2){ListNode head(-1);int carry = 0;ListNode *prev = &head;for(ListNode *pa = l1, *pb = l2;pa != nullptr || pb != nullptr;pa = pa == nullptr ? nullptr : pa->next,pb = pb == nullptr ? nullptr : pb->next,prev = prev->next){const int ai = pa == nullptr ? 0 : pa-val;const int bi = pb == nullptr ? 0 : pb->val;const int value = (ai + bi + carry) % 10;carry = (ai + bi + carry) / 10;prev->next = new ListNode(value);}if(carry > 0)prev->next = new ListNode(carry);return head.next;}
};

参考资料：

LeetCode题解

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