LeetCode之2_Add Two Numbers

题目原文：

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

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分析：输入两个整数序列，将其对应位置上的数字相加（从左到右），注意有进位的情况（原位置只保留其个位数，十位数向右进位），两个序列长度不等的情况和进位在最后新增一个位置的情况。

          

代码：

 

class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        if (l1 == NULL && l2 == NULL)        {return NULL;        }int nAddOverFlow = 0; //相加是否有进位int nTempAdd = 0;int nHeadFlag = 0; ListNode *pRetHead = NULL;ListNode *pTail = NULL;ListNode *pTempPtr = NULL;while (l1!= NULL || l2 != NULL){//若还有节点没有加完，则继续相加，且考虑进位的情况nTempAdd = 0;nTempAdd += nAddOverFlow;if (l1!= NULL){nTempAdd += l1->val;l1 = l1->next;}if (l2!= NULL){nTempAdd += l2->val;l2 = l2->next;}nAddOverFlow = nTempAdd/10;nTempAdd = nTempAdd%10;pTempPtr = (ListNode*)malloc(sizeof(ListNode));pTempPtr->next = NULL;if (nHeadFlag == 0){nHeadFlag = 1;pRetHead = pTempPtr;pTail = pTempPtr;}else{pTail->next = pTempPtr;pTail = pTempPtr;}pTempPtr->val = nTempAdd;}if (nAddOverFlow > 0){pTempPtr = (ListNode*)malloc(sizeof(ListNode));pTempPtr->next = NULL;pTail->next = pTempPtr;pTail = pTempPtr;pTempPtr->val = nAddOverFlow;}return pRetHead;    }};