【Leet Code】Add Two Numbers

来源：互联网发布：ueditor internal.js 编辑：程序博客网时间：2024/04/29 17:58

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

这道题目主要考察对创建链表的运用而已，注意：表头是个位：

/** * 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;    }};

再附上大神的代码：

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;    }};

0 0