Leetcode 2 两数相加

一、题目

　　给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的，并且每个节点只能存储一 数字。请你将两个数相加，并以相同形式返回一个表示和的链表。

　　可以假设除了数字 0 之外，这两个数都不会以 0 开头。

　　示例 1：

　　　　输入：l1 = [2,4,3], l2 = [5,6,4]
　　　　输出：[7,0,8]
　　　　解释：342 + 465 = 807.

二、解法

　　以逆序链表的方式存储数字，非常适合于两数相加。这时候，链表头代表数字的最低位，只需从头向尾遍历，依次相加进位即可。

       以最短的数的链表为终点，然后再将较长的数的后续位加入新的链表中。

　　Python版：

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class Solution:     def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:         upper_val = 0         dummy = ListNode(0)         cur = dummy         while l1 and l2:             value = l1.val + l2.val + upper_val             upper_val = value // 10             tmp_node = ListNode(value % 10)             cur.next = tmp_node             cur = tmp_node             l1, l2 = l1.next, l2.next           remain_l = l1 if l1 else l2         while remain_l:             value = remain_l.val + upper_val             upper_val = value // 10             tmp_node = ListNode(value % 10)             cur.next = tmp_node             cur = tmp_node             remain_l = remain_l.next           if upper_val > 0:             tmp_node = ListNode(upper_val)             cur.next = tmp_node           return dummy.next

       C语言版：

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# include <stdlib.h>   struct ListNode {     int val;     struct ListNode *next; };   struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){     struct ListNode* l = NULL;     struct ListNode* cur = NULL;     // struct ListNode* tmp = NULL;     l = malloc(sizeof(struct ListNode));     l->val = 0;     l->next = NULL;     cur = l;       int value = 0;     int upperVal = 0;     while (l1 != NULL && l2 != NULL) {         value = l1->val + l2->val + upperVal;         upperVal = value / 10;         value = value % 10;                   struct ListNode* tmp_node = malloc(sizeof(struct ListNode));         tmp_node->val = value;         tmp_node->next = NULL;         cur->next = tmp_node;         cur = cur->next;           l1 = l1->next;         l2 = l2->next;     }       if (l1 == NULL) {         while (l2 != NULL ) {             value = l2->val + upperVal;             upperVal = value / 10;             l2->val = value % 10;             cur->next = l2;             cur = cur->next;             l2 = l2->next;         }     }     if (l2 == NULL) {         while (l1 != NULL ) {             value = l1->val + upperVal;             upperVal = value / 10;             l1->val = value % 10;             cur->next = l1;             cur = cur->next;             l1 = l1->next;         }     }     if (upperVal > 0) {         struct ListNode* node = malloc(sizeof(struct ListNode));         node->val = upperVal;         node->next = NULL;         cur->next = node;     }     return l->next; }