LeetCode 002 Add Two Numbers

题目描述：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

注意事项：

1. (2 -> 4 -> 3)中，2是最低位，3是最高位；

2. 两个list的长度不一样；

3. 进位，(9 -> 9) + (1) = (0 -> 0 -> 1)；

4. 不用考虑负数的情况。

解题思路：

1. 定义一个ListNode用来存放结果（result）；

2. 相同位相加，将计算结果的个位传给result，进位传给carry。指针分别后移，如果不为NULL则循环；

3. 最后再次判断进位carry；

4. 返回结果。

代码如下（参考网上）：

/**
 * 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 carry = 0;  //进位标志
        ListNode* result = new ListNode(0);
        ListNode* ptr = result;
        
        while(l1 != NULL || l2 != NULL){
            
            int val1 = 0;
            if(l1 != NULL){
                val1 = l1->val;
                l1 = l1->next;
            }
            
            int val2 = 0;
            if(l2 != NULL){
                val2 = l2->val;
                l2 = l2->next;
            }
            
            int tmp = val1 + val2 + carry;
            ptr->next = new ListNode(tmp % 10);
            carry = tmp / 10;
            ptr = ptr->next;
        }
        
        if(carry == 1){
            ptr->next = new ListNode(1);
        }
        
        return result->next;
        
    }
};

/**
 * 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) {
        
        ListNode* dummyHead = new ListNode(0);
        ListNode* p = l1;
        ListNode* q = l2;
        ListNode* curr = dummyHead;
        
        int carry = 0;
        
        while(p != NULL || q != NULL){
            int x = (p != NULL) ? p->val : 0;
            int y = (q != NULL) ? q->val : 0;
            int digit = carry + x + y;
            carry = digit / 10;
            
            curr->next = new ListNode(digit % 10);
            curr = curr->next;
            
            if(p != NULL) p = p->next;
            if(q != NULL) q = q->next;
        }
        
        if(carry > 0) curr->next = new ListNode(carry);
        
        return dummyHead->next;
    }
};

 Java:

    public class ListNode {

        int val;
        ListNode next;

        ListNode(int x) {
            val = x;
        }
    }

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        int carry = 0;
        ListNode result = new ListNode(0);
        ListNode ptr = result;

        while (l1 != null || l2 != null) {

            int val1 = 0;
            if (l1 != null) {
                val1 = l1.val;
                l1 = l1.next;
            }

            int val2 = 0;
            if (l2 != null) {
                val2 = l2.val;
                l2 = l2.next;
            }

            // 处理相加结果，注意进位
            int tmp = val1 + val2 + carry;
            ptr.next = new ListNode(tmp % 10);
            carry = tmp / 10;
            ptr = ptr.next;
        }

        // 如果最终还有进位，则要新建一个结点！
        if (carry == 1) {
            ptr.next = new ListNode(carry);
        }

        return result.next;
    }