LeetCode Add Two Numbers两个链表数值相加

Add Two Numbers

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

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;
	}