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Add Two Numbers

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Description

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807

Solution

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/**
 * 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) {
		ListNode *tmp1 = l1->next, *tmp2 = l2->next;
		ListNode *res = new ListNode((l1->val + l2->val) % 10);
		ListNode *tmp = res;
		int flag = (l1->val + l2->val) > 9 ? 1 : 0;
		int k;
		while (tmp1&&tmp2)
		{
			k = tmp1->val + tmp2->val+flag;
			tmp->next = new ListNode(k%10);
			flag = k / 10;
			tmp1 = tmp1->next;
			tmp2 = tmp2->next;
			tmp = tmp->next;
		}
		while (tmp1)
		{
			k = tmp1->val+flag;
			tmp->next = new ListNode(k % 10);
			flag = k / 10;
			tmp1 = tmp1->next;
			tmp = tmp->next;
		}
		while (tmp2)
		{
			k = tmp2->val+flag;
			tmp->next = new ListNode(k % 10);
			flag = k / 10;
			tmp2 = tmp2->next;
			tmp = tmp->next;
		}
        if(flag)
        tmp->next = new ListNode(1);
		return res;
	}
};
lessons

题目思路其实就像平时算加法一样，从后向前算，考虑进位。开始傻傻地想要把两个list倒过来，得到两个整数进行相加，然后发现长整形的范围远不能满足题目要求，所以才改成了上面那样。看了看别人的代码，发现我的三个while循环实在有些冗杂，可以整理成一个循环，主要考虑对第一个while循环条件进行修改（ && -> || ），后面的做法就比较类似了。有点乏了，先不改了，就这样吧。