lintcode-【简单题】链表求和

题目：

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中

相反

的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

样例：

给出两个链表

3->1->5->null

和

5->9->2->null

，返回

8->0->8->null

答案：

从头到尾按链表顺序遍历相加就行啦，如果加到最后，进位不为0，还需要另外添加一个节点。

代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2) {
// write your code here
int carry = 0,value = 0;
ListNode *l1Iter = l1;
ListNode *l2Iter = l2;
ListNode *ansRoot = NULL,*ansIter = NULL;

while(l1Iter != NULL && l2Iter != NULL)
{
value = l1Iter->val + l2Iter->val + carry;
carry = value / 10;
value = value % 10;

ListNode *node = new ListNode(value);
if(ansRoot == NULL)
{
ansRoot = node;
}

if(ansIter != NULL)
{
ansIter->next = node;
}

ansIter = node;
l1Iter = l1Iter->next;
l2Iter = l2Iter->next;
}

while(l1Iter != NULL)
{
value = l1Iter->val + carry;
carry = value / 10;
value = value % 10;

ListNode *node = new ListNode(value);
if(ansRoot == NULL)
{
ansRoot = node;
}

if(ansIter != NULL)
{
ansIter->next = node;
}

ansIter = node;
l1Iter = l1Iter->next;
}

while(l2Iter != NULL)
{
value = l2Iter->val + carry;
carry = value / 10;
value = value % 10;

ListNode *node = new ListNode(value);
if(ansRoot == NULL)
{
ansRoot = node;
}

if(ansIter != NULL)
{
ansIter->next = node;
}

ansIter = node;
l2Iter = l2Iter->next;
}

if(carry != 0)
{
ListNode *node = new ListNode(carry);
if(ansIter != NULL)
{
ansIter->next = node;
}
}
return ansRoot;
}
};

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