Leetcode 初探

Add Two Numbers

第一次在leetcode上面做題，選了兩道最簡單的題目來試試手，回憶一下cpp的寫法。寫了一個暑假的python之後，在cpp語法上稍微有點不太習慣，還好，做完前兩道題就上手了。

這裏主要講一講Question2.
題目原意：

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

這種題目以前做的也挺多，就是兩個鏈表相加，注意進位問題。

第一次寫出來的代碼是這樣的（確實有點久沒寫c++,有點生疏…)：

/**
 * 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* ans = new ListNode(l1->val + l2->val);
        ListNode* pre = ans;
        ListNode* cur1 = l1->next, *cur2 = l2->next;
        int carry = 0;
        Carry(ans, carry);
        while (cur1 != NULL || cur2 != NULL) {
            ListNode* node = new ListNode(0);
            if (cur1 != NULL && cur2 != NULL) {
                node->val = cur1->val + cur2->val + carry;
                cur1 = cur1->next;
                cur2 = cur2->next;
            }
            else if (cur1 != NULL) {
                node->val = cur1->val + carry;
                cur1 = cur1->next;
            }
            else if (cur2 != NULL) {
                node->val = cur2->val + carry;
                cur2 = cur2->next;
            }
            pre->next = node;
            pre = node;
            Carry(node, carry);
        }
        if (carry == 1) 
            pre->next = new ListNode(1);
        return ans;
    }
    void Carry(ListNode*& node, int &carry) {
        if (node->val > 9) {
            carry = 1;
            node->val -= 10;
        }
        else 
            carry = 0;
    }
};

看了答案後，發現答案的寫法確實簡潔：

/**
 * 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* dummyHead = new ListNode(0);
        ListNode* p = l1, * q = l2, *cur = dummyHead;
        int carry = 0;
        while (p != NULL || q != NULL) {
            //利用三元表達式處理if語句
            int x = (p != NULL)? p->val: 0;
            int y = (q != NULL)? q->val: 0;
            int sum = x + y + carry;
            //利用除法和取餘處理多種情況
            carry = sum / 10;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if (p != NULL) p = p->next;
            if (q != NULL) q = q->next;
        }
        //最後處理進位
        if (carry)
            cur->next = new ListNode(1);
        return dummyHead->next;
    }
};