前言：
今天我们对“python列表内的数字相加”都比较着重，兄弟们都需要知道一些“python列表内的数字相加”的相关资讯。那么小编也在网摘上搜集了一些关于“python列表内的数字相加””的相关资讯，希望我们能喜欢，大家一起来学习一下吧！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 一位 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。
你可以假设除了数字 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.
示例 2：
输入：l1 = [0], l2 = [0]
输出：[0]
示例 3：
输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]
提示：
每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零
c语言解题：
知识点：单链表的建立，无head，尾插入。
#include<stdio.h>#include<stdlib.h>//Definition for singly-linked list.struct ListNode { int val; struct ListNode *next; }; struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){ int carry = 0; struct ListNode* ret = NULL, *tail = NULL; struct ListNode* l1_head = l1,*l2_head = l2; if (l1_head == NULL && l2_head== NULL) { return ret = NULL; } while (l1_head != NULL || l2_head != NULL) { struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode)); new_node->next = NULL; int l1_head_val = l1_head == NULL ? 0 : l1_head->val; int l2_head_val = l2_head == NULL ? 0 : l2_head->val; new_node -> val = l1_head_val + l2_head_val + carry; carry = (new_node -> val ) / 10 ; new_node -> val =(new_node -> val ) % 10 ; if(l1_head != NULL) l1_head = l1_head ->next; if(l2_head != NULL) l2_head = l2_head ->next; if(ret == NULL) ret = new_node; else { tail->next = new_node; } tail = new_node; } if (carry) { struct ListNode* new_node = (struct ListNode*)malloc(sizeof(struct ListNode)); new_node->next = NULL; new_node -> val = carry; if(ret == NULL) ret = new_node; else { tail->next = new_node; } tail = new_node; } return ret; }int main(){ int l1arr[] = {9, 9, 9, 9, 9, 9, 9}; int l2arr[] = {9, 9, 9, 9}; struct ListNode* l1, *l1tail, *p1; struct ListNode* l2, * l2tail, *p2; struct ListNode* ret; int l1arrSize = sizeof(l1arr) / sizeof(int); int l2arrSize = sizeof(l2arr) / sizeof(int); l1tail = p1 = (struct ListNode*)malloc(sizeof(struct ListNode)); l2tail = p2 = (struct ListNode*)malloc(sizeof(struct ListNode)); l1 = NULL; l2 = NULL; for (int i = 0; i < l1arrSize; i++) { p1 ->val = l1arr[i]; p1 ->next = NULL; if(l1 == NULL) l1 = p1; else { l1tail->next = p1; } l1tail = p1; p1 = (struct ListNode*)malloc(sizeof(struct ListNode)); } l1tail = NULL,p1 = NULL; for (int i = 0; i < l2arrSize; i++) { p2 ->val = l2arr[i]; p2 ->next = NULL; if(l2 == NULL) l2 = p2; else { l2tail->next = p2; } l2tail = p2; p2 = (struct ListNode*)malloc(sizeof(struct ListNode)); } l2tail = NULL,p2 = NULL; ret = addTwoNumbers(l1, l2); printf("[%d,%d,%d,%d,%d,%d,%d,%d]",ret->val,ret->next->val,ret->next->next->val ,ret->next->next->next->val,ret->next->next->next->next->val,ret->next->next->next->next->next->val ,ret->next->next->next->next->next->next->val,ret->next->next->next->next->next->next->next->val); return 0;}python3解题：
# Definition for singly-linked list.class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = nextclass Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: ret = ListNode(0) tail = ret carry = 0 while(l1 or l2): val1 = l1.val if l1 else 0 val2 = l2.val if l2 else 0 sum = carry + val1 + val2 carry = sum // 10 tail.next = ListNode(sum % 10) tail = tail.next if(l1 is not None): l1 = l1.next if(l2 is not None): l2 = l2.next if (carry > 0): tail.next = ListNode(1) return ret.nextdef build_single_link(nums): tail = head = ListNode(None) for i in nums: tail.next = ListNode(i) tail = tail.next return head.nextnums1 = [2, 4, 3]nums2 = [5, 6, 4]l1 = build_single_link(nums1)l2 = build_single_link(nums2)test = Solution()res = test.addTwoNumbers(l1, l2)while res: print(res.val, end=' ') res = res.nextprint()
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标签： #python列表内的数字相加
