Leetcode2：Add Two Numbers

题目描述

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.

分析

本题考查链表的使用，比较简单。并且数字是低位在前，直接计算即可，注意进位的保存。关于链表的题目都比较直接，难度不大。不过链表写起来可能比较繁琐一些，一般需要先创建头结点，这样比较容易找到链表首部。

AC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) {
        int flag = 0, value = 0;//flag保存进位
        value = (l1->val + l2->val + flag) % 10;
        ListNode* temp1 = new ListNode(value);
        flag = (l1->val + l2->val + flag) / 10;
        ListNode* root = temp1;//头结点单独处理
        l1 = l1->next;
        l2 = l2->next;
        while(l1 != NULL && l2 != NULL)//链表是否到尽头
        {
        	value = (l1->val + l2->val + flag) % 10;
        	ListNode* temp2 = new ListNode(value);
        	flag = (l1->val + l2->val + flag) / 10;
        	temp1->next = temp2;
        	temp1 = temp2;
        	l1 = l1->next;
        	l2 = l2->next;
		}
		while(l1 != NULL)//l1还有高位
		{
			value = (l1->val + flag) % 10;
        	ListNode* temp2 = new ListNode(value);
        	flag = (l1->val + flag) / 10;
        	temp1->next = temp2;
        	temp1 = temp2;
        	l1 = l1->next;
		}
		while(l2 != NULL)//l2还有高位
		{
			value = (l2->val + flag) % 10;
        	ListNode* temp2 = new ListNode(value);
        	flag = (l2->val + flag) / 10;
        	temp1->next = temp2;
        	temp1 = temp2;
        	l2 = l2->next;
		}
		if(flag != 0)//如果还有进位，需要进到更高位
		{
        	ListNode* temp2 = new ListNode(flag);
        	temp1->next = temp2;
		}
		return root;//返回头结点
    }
};