LeetCode2 两数相加 两个链表相加(C++)

题目链接

题目描述

给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储一位数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例 1：

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

示例 2：

输入：l1 = [0], l2 = [0]
输出：[0]

示例 3：
输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

提示：
每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零

LeetCode实现（C++）

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {if (l1 == NULL || l2 == NULL){if (l2 == NULL){return l1;}if (l1 == NULL){return l2;}}bool carry = false;	// 进位ListNode* l3 = new ListNode();ListNode* Ret = l3;	// 记录返回链表头结点while (l1 != NULL || l2 != NULL){if (carry){l3->val++;}if (l1){l3->val += l1->val;l1 = l1->next;}if (l2){l3->val += l2->val;l2 = l2->next;}if (l3->val >= 10){carry = true;l3->val -= 10;}else{carry = false;}if (l1 != NULL || l2 != NULL){l3->next = new ListNode();l3 = l3->next;}}if (carry){l3->next = new ListNode(carry);}return Ret;}

完整实现和测试(C++)

#include <iostream>
#include <string>
using namespace std;struct ListNode {int val;ListNode* next;ListNode() : val(0), next(nullptr) {}ListNode(int x) : val(x), next(nullptr) {}ListNode(int x, ListNode* next) : val(x), next(next) {}
};class Solution2 
{public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {if (l1 == NULL || l2 == NULL){if (l2 == NULL){return l1;}if (l1 == NULL){return l2;}}bool carry = false;	// 进位ListNode* l3 = new ListNode();ListNode* Ret = l3;	// 记录返回链表头结点while (l1 != NULL || l2 != NULL){if (carry){l3->val++;}if (l1){l3->val += l1->val;l1 = l1->next;}if (l2){l3->val += l2->val;l2 = l2->next;}if (l3->val >= 10){carry = true;l3->val -= 10;}else{carry = false;}if (l1 != NULL || l2 != NULL){l3->next = new ListNode();l3 = l3->next;}}if (carry){l3->next = new ListNode(carry);}return Ret;}void ShowList(ListNode* L)	// 显示链表中所有值{while (L != NULL){cout << L->val << " ";L = L->next;}cout << endl;}void DeleteList(ListNode* L)	// 清理内存{if(!L){return ;}ListNode* p = L->next;	// 先保存下一个结点delete L;L = p;while (L != NULL){p = L->next;	// 先保存delete L;L = p;}}ListNode* InitList(string strNum){if (strNum.empty()){return NULL;}ListNode* tmp = new ListNode();ListNode* Ret = tmp;for (int i = 0; i < strNum.length(); i++){tmp->val = atoi(strNum.substr(i, 1).c_str());if (i + 1 < strNum.length()){tmp->next = new ListNode();}tmp = tmp->next;}return Ret;}
};void test(string str1, string str2)
{ListNode* l1;ListNode* l2;ListNode* l3;Solution2 obj;cout << "--------------------- str1:" << str1 << " str2:" << str2 << " ---------------------" << endl;l1 = obj.InitList(str1);l2 = obj.InitList(str2);l3 = obj.addTwoNumbers(l1, l2);obj.ShowList(l1);obj.ShowList(l2);obj.ShowList(l3);obj.DeleteList(l1);obj.DeleteList(l2);obj.DeleteList(l3);cout << "---------------------------------------------------------------\n" << endl;
}int main()
{// 输入：l1 = [2, 4, 3], l2 = [5, 6, 4]// 输出：[7, 0, 8]test("243", "564");// 输入：l1 = [0], l2 = [0]// 输出：[0]test("0", "0");// 输入：l1 = [9, 9, 9, 9, 9, 9, 9], l2 = [9, 9, 9, 9]// 输出：[8, 9, 9, 9, 0, 0, 0, 1]test("9999999", "9999");test("1", "9999");test("99", "9");return 0;
}