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2. 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
题目分析：
给定两个链表（代表两个非负数），数字的各位以倒序存储，将两个代表数字的链表想加获得一个新的链表（代表两数之和）。
如(2->4->3)(342) + (5->6->4)(465) = (7->0->8)(807)
设两个进行加法运算的链表分别为l1,l2, sum链表为l3.
若以l[i] 表示链表各个节点的值，nCarryBit[i]表示l[i]位相加产生的进位符
则有以下结论:
l3[i] = (l1[i] + l2[i] + nCarryBit[i-1]) % 10
nCarryBit[i] = (l1[i] + l2[i] + nCarryBit[i-1]) / 10
且nCarryBit[0] = 0;
Solution
/** * 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 v1,v2; int nCarryBit = 0; ListNode* Prehead = new ListNode(0); ListNode* l3 = Prehead; while(l1 || l2 ||nCarryBit) { v1 = 0, v2 = 0; if(l1) { v1 = l1->val; l1 = l1->next; } if(l2) { v2 = l2->val; l2 = l2->next; } int sum = v1 + v2 +nCarryBit; nCarryBit = sum/10; l3->next = new ListNode(sum%10); l3 = l3->next; } return Prehead->next; } };
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原文地址：http://www.cnblogs.com/HellcNQB/p/5370886.html
