题目:两数相加
- 题号:2
- 难度:中等
- https://leetcode-cn.com/problems/add-two-numbers/
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
实现
思路:模仿我们小学时代学的加法运算。个位相加超过十进一,十位相加有进位则加上进位,依次类推。
C# 实现
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution
{
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
ListNode result = new ListNode(-1);
ListNode l3 = result;
int flag = 0;
while (l1 != null && l2 != null)
{
int a = l1.val;
int b = l2.val;
int c = a + b + flag;
l3.next = new ListNode(c%10);
flag = c >= 10 ? 1 : 0;
l1 = l1.next;
l2 = l2.next;
l3 = l3.next;
}
while (l1 != null)
{
int a = l1.val + flag;
l3.next = new ListNode(a%10);
flag = a >= 10 ? 1 : 0;
l1 = l1.next;
l3 = l3.next;
}
while (l2 != null)
{
int b = l2.val + flag;
l3.next = new ListNode(b%10);
flag = b >= 10 ? 1 : 0;
l2 = l2.next;
l3 = l3.next;
}
if (flag == 1)
{
l3.next = new ListNode(flag);
}
return result.next;
}
}
Python 实现
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
result = ListNode(-1)
l3 = result
flag = 0
while l1 is not None and l2 is not None:
a = l1.val
b = l2.val
c = a + b + flag
l3.next = ListNode(c % 10)
flag = 1 if c >= 10 else 0
l1 = l1.next
l2 = l2.next
l3 = l3.next
while l1 is not None:
a = l1.val + flag
l3.next = ListNode(a % 10)
flag = 1 if a >= 10 else 0
l1 = l1.next
l3 = l3.next
while l2 is not None:
b = l2.val + flag
l3.next = ListNode(b % 10)
flag = 1 if b >= 10 else 0
l2 = l2.next
l3 = l3.next
if flag == 1:
l3.next = ListNode(flag)
return result.next