leetcode_add_two_numbers_ii

发表于

难度：Medium

解题思路：因为链表从前到后，是由高到低位。所以这里进行了翻转，然后计算。将链表之和放在了l1上，如果l2的长度超出了l1，则将超出的部分，接到l1的尾部。

代码如下：

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/**
 * 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) {
        l1 = reverseLinkedList(l1);
        l2 = reverseLinkedList(l2);
        ListNode *l1_cur = l1;
        ListNode *l2_cur = l2;
        ListNode *l1_cur_pre = NULL;
        ListNode *l2_cur_pre = NULL;
        int add = 0;
        while(l1_cur != NULL && l2_cur != NULL)
        {
            int sum = add + l1_cur->val + l2_cur->val;
            add = (sum >= 10)?1:0;
            // cout<<add<<endl;
            sum = sum % 10;
            l1_cur->val = sum;
            l1_cur_pre = l1_cur;
            l2_cur_pre = l2_cur;
            l1_cur = l1_cur->next;
            l2_cur = l2_cur->next;
        }
        if(l2_cur != NULL)
        {
            l1_cur_pre->next = l2_cur;
            l2_cur_pre->next = NULL;
            l1_cur = l1_cur_pre->next;
        }
        while(l1_cur != NULL)
        {
            int sum = add + l1_cur->val;
            add = (sum >= 10)?1:0;
            sum = sum % 10;
            l1_cur->val = sum;  
            l1_cur_pre = l1_cur;
            l1_cur = l1_cur->next;
        }
        if(add == 1)
        {
            ListNode* new_node = new ListNode(add);
            l1_cur_pre->next = new_node;
        }
        l1 = reverseLinkedList(l1);
        return l1;
    }
    ListNode* reverseLinkedList(ListNode* head)
    {
        ListNode *cur = head;
        ListNode *pre = NULL;
        while(cur != NULL)
        {
            ListNode *ne = cur->next;
            cur->next = pre;
            pre = cur;
            cur = ne;
        }
        return pre;
    }
};

运行结果：35ms，超过61.02%