leetcode-個人題解2

LEETCODE專題

---

Add Two Numbers

首先先上題目：

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

---

下面是一些個人理解和細節把握：

1. 鏈表的每個節點其實都可以相加然後往下走到下一個節點，但是要把握好前提條件：

   1. 兩個鏈表要保持同一個深度
   2. 每一個節點相加後的進位要存儲至結果鏈表的下一個節點

   這些都是可以做到的，下面有代碼，在這裏不再贅述

2.在1[1]的基礎上提出問題：倘若兩個鏈表不等長怎麼辦？
這時候就加一個條件判斷就能夠解決問題

3.最後倘若兩個輸入鏈表已經結束然而進位爲1，則要在結果鏈表的尾端新建一個結果節點，將進位存進去

---

下面直接上代碼：

/**
 * 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) {
        bool first = true;                    // firstly enter the while block
        int carry = 0;                        // carry = 1 or 0
        ListNode * sum = new ListNode(0);     // store the sum of l1 and l2
        ListNode * head_of_sum = sum;         // return the head of ListNode * sum
        while (l1 != NULL || l2 != NULL) {
            int temp_sum = carry;

            if (first == false) {
                sum->next = new ListNode(0);
                sum = sum->next;
            } else {
                first = false;
            }

            if (l1 == NULL) {
                temp_sum += l2->val;
            } else if (l2 == NULL) {
                temp_sum += l1->val;
            } else {
                temp_sum += l1->val + l2->val;
            }

            carry = temp_sum / 10;
            sum->val = temp_sum % 10;

            if (l1 != NULL) {
                l1 = l1->next;
            }
            if (l2 != NULL) {
                l2 = l2->next;
            }
        }

        if (carry == 1) {
            sum->next = new ListNode(carry);
        }
        return head_of_sum;
    }
};