lintcode-【簡單題】鏈表求和

題目：

你有兩個用鏈表代表的整數，其中每個節點包含一個數字。數字存儲按照在原來整數中相反的順序，使得第一個數字位於鏈表的開頭。寫出一個函數將兩個整數相加，用鏈表形式返回和。

樣例：

給出兩個鏈表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

答案：

從頭到尾按鏈表順序遍歷相加就行啦，如果加到最后，進位不為0，還需要另外添加一個節點。

代碼：

/**
 * 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;
    }
};