leetcode——2——Add Two Numbers

来源：互联网发布：家用网络如何接交换机编辑：程序博客网时间：2024/06/05 01:52

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

要注意边做进位加法边创建链表，一开始一直用node=node->next;这样链表不存在，不能得出结果，并考虑到最后一位还有进位的情况

（ps：第一次做leetcode，如果有时间还要回来看看，刷了10多天终于做到Medium了，要加快进度）

/** * 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 = (ListNode *)malloc(sizeof(ListNode));        ListNode *pre = head;        ListNode *node = NULL;        //进位        int c = 0,sum;        //加法        while(l1 != NULL && l2 != NULL){            sum = l1->val + l2->val + c;            c = sum / 10;            node = (ListNode *)malloc(sizeof(ListNode));            node->val = sum % 10;            node->next = NULL;            //尾插法            pre->next = node;            pre = node;            l1 = l1->next;            l2 = l2->next;        }        //例如：2->4->3->1   5->6->4        while(l1 != NULL){            sum = l1->val + c;            c = sum / 10;            node = (ListNode *)malloc(sizeof(ListNode));            node->val = sum % 10;            node->next = NULL;            //尾插法            pre->next = node;            pre = node;                       l1 = l1->next;        }        //例如：2->4->3   5->6->4->1        while(l2 != NULL){            sum = l2->val + c;            c = sum / 10;            node = (ListNode *)malloc(sizeof(ListNode));            node->val = sum % 10;            node->next = NULL;            //尾插法            pre->next = node;            pre = node;            l2 = l2->next;        }        //最后一位还有进位        if(c > 0){            node = (ListNode *)malloc(sizeof(ListNode));            node->val = c;            node->next = NULL;            //尾插法            pre->next = node;            pre = node;        }        return head->next;            }};

0 0