2. Add Two Numbers

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https://leetcode.com/problems/add-two-numbers/description/

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先將 l1, l2 加完，接著把剩下的 list 跟 carry 加完即可

/**
 * Definition for singly-linked list.
 * 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 Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(); ListNode* cur = head;
        int carry = 0;
        while(l1 && l2) {
            int sum = l1->val + l2->val + carry;
            cur->next = new ListNode(sum % 10);
            cur = cur->next; l1 = l1->next; l2 = l2->next;
            carry = sum / 10;
        }
        ListNode* remain = l1 ? l1 : l2;
        while(carry || remain) {
            if(!remain) {cur->next = new ListNode(carry); break;}
            int sum = remain->val + carry;
            cur->next = new ListNode(sum % 10);
            cur = cur->next; remain = remain->next;
            carry = sum / 10;
        }
        return head->next;
    }
};

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簡化寫法，走到盡頭的視作 0 不貢獻值，判斷 l1, l2, carry 是否還有值，繼續運算

/**
 * Definition for singly-linked list.
 * 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 Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(); ListNode* cur = head;
        int carry = 0;
        while(l1 || l2 || carry) {
            int a = l1 ? l1->val : 0;
            int b = l2 ? l2->val : 0;
            int sum = a + b + carry;
            cur->next = new ListNode(sum % 10);
            carry = sum / 10;
            cur = cur->next; 
            l1 = l1 ? l1->next : l1;
            l2 = l2 ? l2->next : l2;
        }
        return head->next;
    }
};