• [LeetCode] Add Two Numbers


    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

    思路:两条链表相加,注意最后如果进位不为零,要加入到结果中。使用二级指针记录节点前缀。

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    12         if (l1 == NULL || l2 == NULL)
    13             return l1 == NULL ? l2 : l1;
    14             
    15         int carry = 0;
    16         ListNode *pter1 = l1;
    17         ListNode *pter2 = l2;
    18         ListNode *head = NULL;
    19         ListNode **prev= &head;
    20         while (pter1 != NULL || pter2 != NULL) {
    21             const int l1_val = pter1 == NULL ? 0 : pter1->val;
    22             const int l2_val = pter2 == NULL ? 0 : pter2->val;
    23             int temp_val = l1_val + l2_val + carry;
    24             ListNode *temp_node = new ListNode(temp_val % 10);
    25             carry = temp_val / 10;
    26             *prev = temp_node;
    27             prev = &(temp_node->next);
    28             pter1 = pter1 == NULL ? NULL : pter1->next;
    29             pter2 = pter2 == NULL ? NULL : pter2->next;
    30         }
    31         
    32         if (carry == 1) {
    33             ListNode *temp_node = new ListNode(1);
    34             *prev = temp_node;
    35         }
    36         
    37         return head;
    38     }
    39 };
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  • 原文地址:https://www.cnblogs.com/vincently/p/4055340.html
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