LeetCode 2. Add Two Numbers--C++,Python解法--面试算法题

题目地址：Add Two Numbers - LeetCode

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

最容易想到的解法是把链表转成数字相加后再转成链表。
Python解法如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:num1, num2 = '', ''while l1 != None:num1 = num1+str(l1.val)l1 = l1.nextwhile l2 != None:num2 = num2+str(l2.val)l2 = l2.nextnum = int(int(num1[::-1])+int(num2[::-1]))root = ListNode(x=0)temp2 = rootwhile num != 0:temp = num % 10temp2.val = tempnum = num//10if num == 0:return roottemp3 = ListNode(x=0)temp2.next = temp3temp2 = temp3return root

这个做法有缺点，那就是如果数很大的话会比较麻烦，虽然Python的int可以无限大，但C语言就不能用这个解法了。
比较好的做法如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:dummy = cur = ListNode(0)carry = 0while l1 or l2 or carry:if l1:carry += l1.vall1 = l1.nextif l2:carry += l2.vall2 = l2.nextcur.next = ListNode(carry%10)cur = cur.nextcarry //= 10return dummy.next

C++解法如下：

/*** 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){struct ListNode *ans = NULL;struct ListNode *change, *az;int temp = 0;while (l1 || l2 || temp){if (l1){temp += l1->val;l1 = l1->next;}if (l2){temp += l2->val;l2 = l2->next;}az = new struct ListNode(0);az->next = NULL;az->val = temp % 10;if (ans == NULL){ans = az;change = ans;}else{change->next = az;change = change->next;}temp = temp / 10;}return ans;}
};

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