[Leetcode] 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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

题目大意：链表中，每个节点表示多位数中的某一位，每个节点的value一定是单个数字，把两个链表的值相加得到一个新的链表

首先是原创的方法，代码很乱，读者可以忽视这部分，看后一部分，因为自己不会创建链表，因此返回的是输入中的某一个链表

方法：找到更长的链表，并且每次加法后覆盖其原有的value，最后针对超出的部分进行处理即可

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int c = 0;
        ListNode* root_l1 = l1, *root_l2 = l2;
        int len_l1 = 0, len_l2 = 0;
        while(root_l1 != NULL) {++len_l1; root_l1 = root_l1->next;}
        while(root_l2 != NULL) {++len_l2; root_l2 = root_l2->next;}
        int length = len_l1 > len_l2 ? len_l2 : len_l1;
        ListNode* root = len_l1 > len_l2 ? l1 : l2;
        ListNode* head = new ListNode(-1);
        head->next = root; 
        for(int i = 0; i < length; ++i){
            int tmp = (l1->val + l2->val + c) / 10;
            head->next->val = (l1->val + l2->val + c) % 10;
            c = tmp;
            head = head->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(head->next != NULL && c != 0){
            int tmp = (head->next->val + c) / 10;
            head->next->val = (head->next->val + c) % 10;
            c = tmp;
            head = head->next;
        }
        
        ListNode* back = new ListNode(1);
        if(head->next == NULL && c == 1){
            head->next = back;
        }
        return root;
    }
};

接下来是参考了网友的方法，代码简洁很多，思路和上面一样，代码优化的地方

1. 创建链表的方式：head->next = new ListNode((l1_val + l2_val + c) % 10);
2. 通过的||+循环内部if的方式，将长链表超出部分的处理也写在一个循环里面

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int c = 0;
        ListNode* head = new ListNode(-1);
        ListNode* root = head;
        while(l1 || l2){
            int l1_val = l1 != NULL ? l1->val : 0;
            int l2_val = l2 != NULL ? l2->val : 0;
            head->next = new ListNode((l1_val + l2_val + c) % 10);
            c = (l1_val + l2_val + c) / 10;
            head = head->next;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }
        if(c){
            head->next = new ListNode(1);
        }
        return root->next;
    }
};

读者有何问题，可以留言沟通，?