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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
这一题主要是熟悉单链表的操作，链表操作使用一个pHead记录链表头位置，还需要使用一个不停向后移动的指针pMove来遍历链表中的元素。
我第一次的思路及代码如下：
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { int iCarry = 0; ListNode *pL1 = l1; ListNode *pL2 = l2; ListNode *pRes = NULL; ListNode *pMove = NULL; ListNode * pTemp = NULL; int flag = 1; while(pL1 != NULL && pL2 != NULL) { pTemp = new ListNode(sizeof(struct ListNode)); if(flag == 1) { pMove = pRes = pTemp; ++flag; } else { pMove->next = pTemp; pMove = pTemp; } pTemp->val = pL1->val + pL2->val + iCarry; if(pTemp->val >= 10) { pTemp->val = pTemp->val % 10; iCarry = 1; } else { iCarry = 0; } pL1 = pL1->next; pL2 = pL2->next; } while(pL1 != NULL || pL2 != NULL) { if( pL2 != NULL ) { pTemp = new ListNode (sizeof(struct ListNode)); pMove->next = pTemp; pMove = pTemp; pMove->val = pL2->val + iCarry; iCarry = 0; if(pMove->val >= 10) { pMove->val %= 10; iCarry = 1; pL2= pL2->next; } else { pL2= pL2->next; } } if(pL1 != NULL) { pTemp = new ListNode (sizeof(struct ListNode)); pMove->next = pTemp; pMove = pTemp; pMove->val = pL1->val + iCarry; iCarry = 0; if(pMove->val >= 10) { pMove->val %= 10; iCarry = 1; pL1= pL1->next; } else { pL1= pL1->next; } } } if(pL1 == NULL && pL2 == NULL && iCarry != 0) { pTemp = new ListNode (sizeof(struct ListNode)); pMove->next = pTemp; pMove = pTemp; pTemp->val = 1; } return pRes; }
真是一段又长又臭的代码，这差不多也就是我现在的水平，还亟待提高。逻辑关系没有理很清楚，代码没有做到最优。
仔细想，其实这里的加法，只需用四个变量iVal1,iVal2,iCarry,iRes来分别记录数据处理的过程，然后就是依次存储到链表中。
就是这段：
if(flag == 1)
{
pMove = pRes = pTemp;
++flag;
}
else
{
pMove->next = pTemp;
pMove = pTemp;
}
后来参考网上代码，所有情况其实只需考虑l1 l2是否非空即可，如下。
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
{
int iCarry = 0;
int iFlag = 1;
ListNode *pTemp = NULL;
ListNode *pHead = NULL;
ListNode *pMove = NULL;
while(l1 != NULL || l2 != NULL)
{
int iVal1 = 0;
if(l1 != NULL)
{
iVal1 = l1->val;
l1 = l1->next;
}
int iVal2 = 0;
if(l2 != NULL)
{
iVal2 = l2->val;
l2 = l2->next;
}
int temp = iVal1 + iVal2 + iCarry;
pTemp = new ListNode(sizeof(ListNode));
if(iFlag == 1)
{
pHead = pMove = pTemp;
++iFlag;
}
else
{
pMove->next = pTemp;
pMove = pTemp;
}
pTemp->val = temp % 10;
iCarry = temp / 10;
}
if(iCarry == 1)
{
pTemp = new ListNode(sizeof(ListNode));
pMove->next = pTemp;
pMove = pTemp;
pTemp->val = 1;
}
return pHead;
}
逻辑关系明确，这才是最终的best code。相信自己最终也能达到这样的水平。
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原文地址：http://www.cnblogs.com/bestwangjie/p/4356089.html
