LeetCode - Add Two Numbers

题目：
/*
给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
*/

解决方案如下：
（以下代码包含了测试代码部分）

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr) {}

    // 为了方便测试，加入该方法，将数字转换成node
    static ListNode *nodeWithNumber(int num) {
        if (num == 0)
        {
            return new ListNode(0);
        }
        int rest = num / 10;
        ListNode *head = new ListNode(num % 10);
        ListNode *tail = head;
        while (rest > 0)
        {
            ListNode *next = new ListNode(rest % 10);
            tail->next = next;
            tail = next;
            rest = rest / 10;
        }
        return head;
    }
};

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if (l1 == nullptr)
        {
            return l2;
        }
        if (l2 == nullptr)
        {
            return l1;
        }

        int sum = 0;
        ListNode *head = new ListNode(0);
        ListNode *cur = head;

        while (1)
        {
            if (l1 != nullptr)
            {
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2 != nullptr)
            {
                sum += l2->val;
                l2 = l2->next;
            }
            cur->val = sum % 10;
            sum = sum / 10;
            if (l1 != nullptr || l2 != nullptr || sum)
            {
                cur->next = new ListNode(0);
                cur = cur->next;
            }
            else
                break;
        }

        return head;
    }
};

void testNode();
void printNode(ListNode *node, bool result);
void printNode(ListNode *node);
void testNormal();
void testSpec();


#include "iostream"
int main() {
    
    testNode(); 
    testNormal();
    testSpec();

    system("pause");

    return 0;
}

void testNode(){
    ListNode *node1 = ListNode::nodeWithNumber(342);
    ListNode *node2 = ListNode::nodeWithNumber(0);
    printNode(node1);
    printNode(node2);
    printf("-----------------------------\n");

    free(node1);
    free(node2);
}

void testNormal() {
    ListNode *node1 = ListNode::nodeWithNumber(342);
    ListNode *node2 = ListNode::nodeWithNumber(465);
    printNode(node1);
    printNode(node2);
    ListNode *result1 = Solution().addTwoNumbers(node1, node2);
    printNode(result1, true);
    printf("-----------------------------\n");

    ListNode *node3 = ListNode::nodeWithNumber(32);
    ListNode *node4 = ListNode::nodeWithNumber(465);
    printNode(node3);
    printNode(node4);
    ListNode *result2 = Solution().addTwoNumbers(node3, node4);
    printNode(result2, true);
    printf("-----------------------------\n");

    free(node1);
    free(node2);
    free(node3);
    free(node4);
    free(result1);
    free(result2);
}

void testSpec() {
    ListNode *node1 = ListNode::nodeWithNumber(0);
    ListNode *node2 = ListNode::nodeWithNumber(465);
    printNode(node1);
    printNode(node2);
    ListNode *result1 = Solution().addTwoNumbers(node1, node2);
    printNode(result1, true);
    printf("-----------------------------\n");

    ListNode *node3 = ListNode::nodeWithNumber(342);
    ListNode *node4 = ListNode::nodeWithNumber(0);
    printNode(node3);
    printNode(node4);
    ListNode *result2 = Solution().addTwoNumbers(node3, node4);
    printNode(result2, true);
    printf("-----------------------------\n");

    ListNode *result3 = Solution().addTwoNumbers(nullptr, node4);
    printNode(result3, true);
    printf("-----------------------------\n");

    ListNode *result4 = Solution().addTwoNumbers(nullptr, nullptr);
    printNode(result4, true);
    printf("-----------------------------\n");

    free(node1);
    free(node2);
    free(node3);
    free(node4);
    free(result1);
    free(result2);
// 因为这里result3得到的就是node4，所以不需要释放result3
// result = nullptr
}


void printNode(ListNode *node) {
    printNode(node, false);
}

void printNode(ListNode *node, bool result) {
    if (node != nullptr)
    {
        ListNode *p = node;
        int num = 0;
        int i = 0;
        while (p != nullptr)
        {
            if (i == 0)
            {
                num += p->val;
            }
            else
            {
                num += p->val * (int)pow(10, i);
            }
            i++;
            printf("%d -> ", p->val);
            p = p->next;
        }
        
        printf("NULL <<<< %s == %d\n", result?"result":"num", num);
    }
    else
    {
        printf("node is a nullptr!\n");
    }
}

执行结果如下：

2 -> 4 -> 3 -> NULL <<<< result == 342
0 -> NULL <<<< result == 0
-----------------------------
2 -> 4 -> 3 -> NULL <<<< result == 342
5 -> 6 -> 4 -> NULL <<<< result == 465
7 -> 0 -> 8 -> NULL <<<< result == 807
-----------------------------
2 -> 3 -> NULL <<<< result == 32
5 -> 6 -> 4 -> NULL <<<< result == 465
7 -> 9 -> 4 -> NULL <<<< result == 497
-----------------------------
0 -> NULL <<<< result == 0
5 -> 6 -> 4 -> NULL <<<< result == 465
5 -> 6 -> 4 -> NULL <<<< result == 465
-----------------------------
2 -> 4 -> 3 -> NULL <<<< result == 342
0 -> NULL <<<< result == 0
2 -> 4 -> 3 -> NULL <<<< result == 342
-----------------------------
0 -> NULL <<<< result == 0
-----------------------------
node is a nullptr!
-----------------------------
请按任意键继续. . .