【Leet Code】Add Two Numbers

Add Two Numbers

 Total Accepted: 20255 Total Submissions: 88115My Submissions

You are given two linked lists representing two non-negative numbers. 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.

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

這道題目主要考察對創建鏈表的運用而已，注意：表頭是個位：

/**
 * 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* head = NULL;
        ListNode* cur = NULL;
        int ad = 0;
        while(l1 && l2)
        {
            if(NULL ==head)
            {
                head = cur = new ListNode( (l1->val + l2->val + ad) % 10 );
            }
            else
            {
                cur->next = new ListNode( (l1->val + l2->val + ad) % 10 );
                cur = cur->next;
            }
            ad = (l1->val + l2->val + ad ) / 10;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1)
        {
            cur->next = new ListNode( (l1->val + ad) % 10 );
            ad = (l1->val + ad) / 10;
            cur = cur->next;
            l1 = l1->next;
        }
        while(l2)
        {
            cur->next = new ListNode( (l2->val + ad) % 10 );
            ad = (l2->val + ad) / 10;
            cur = cur->next;
            l2 = l2->next;
        }
        if (ad > 0) 
        {
            cur->next = new ListNode(ad);
        }
        return head;
    }
};

再附上大神的代碼：

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *ret = NULL;
        ListNode **pCur = &ret;
        int nxt = 0;
        while (l1 && l2) {
            *pCur = new ListNode((l1->val + l2->val + nxt) % 10);
            nxt = (l1->val + l2->val + nxt) / 10;
            pCur = &((*pCur)->next);
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1 != NULL) {
            *pCur = new ListNode((l1->val + nxt) % 10);
            nxt = (l1->val + nxt) / 10;
            pCur = &((*pCur)->next);
            l1 = l1->next;
        }
        while (l2 != NULL) {
            *pCur = new ListNode((l2->val + nxt) % 10);
            nxt = (l2->val + nxt) / 10;
            pCur = &((*pCur)->next);
            l2 = l2->next;
        }
        if (nxt > 0) {
            *pCur = new ListNode(nxt);
        }
        return ret;
    }
};