LeetCode 2. Add Two Numbers

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

题目描述：
给出两个数字链表，求出这两个数字的和并且以类似的链表返回。这里的链表里面的数字顺序是倒序的，也就是说第一个是个位，第二个是十位，第三个是百位……
题目给出了这个链表的结构如下：

//Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};

解题思路：
我的做法是直接将两个链表的数字逐位相加，并且保存两个数字相加后的进位情况，如果相加后的数是一个两位数，那么要将这个数对10取模保留个位数的值，然后把进位标志设为1，在下一位相加的时候把这个进位也加上去。

代码：

#include <iostream>#include <string>using namespace std;//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* result = NULL;        ListNode* temp = NULL;        int carry = 0;        int current = 0;        while (l1 != NULL || l2 != NULL) {            current = 0;            if (l1 != NULL) {                current += l1->val;                l1 = l1->next;            }            if (l2 != NULL) {                current += l2->val;                l2 = l2->next;            }            current += carry;            carry = current / 10;            current %= 10;            if (result == NULL) {                result = new ListNode(current);                temp = result;            }            else {                result->next = new ListNode(current);                result = result->next;            }            if (l1 == NULL && l2 == NULL && carry != 0) {                result->next = new ListNode(carry);            }        }        return temp;    }};void PrintList(ListNode* l) {    if (l == NULL) return;    ListNode* currentNode = l;    while (currentNode != NULL) {        if (currentNode != l) {            cout << " -> ";        }        cout << currentNode->val;        currentNode = currentNode->next;    }}ListNode* Convert2ListNode(string val) {    if (val == "") return NULL;    ListNode* l = new ListNode(val[val.length() - 1] - '0');    ListNode* start = l;    std::string::reverse_iterator r_it = val.rbegin() + 1;    for (; r_it != val.rend(); r_it++) {        l->next = new ListNode(*r_it - '0');        l = l->next;    }    return start;}int main() {    Solution sln;    ListNode* l1 = Convert2ListNode("5");    ListNode* l2 = Convert2ListNode("5");    PrintList(l1);    cout << endl;    PrintList(l2);    cout << endl;    ListNode* l3 = sln.addTwoNumbers(l1, l2);    PrintList(l3);    return 0;}