蜗牛慢慢爬 LeetCode 2. Add Two Numbers [Difficulty: Medium]
题目
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
翻译
给定两个非空的链表,表示两个非负整数。数字倒序存储,每个节点包含一个数字。将两个数字求和并将结果以链表形式返回
你可以假定数字不包含前导0(除了0本身之外)
输入:(2 - > 4 - > 3)+(5 - > 6 - > 4)
输出:7 - > 0 - > 8
Hints
Related Topics: Linked List, Math
注意简化代码
代码
Java
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode c1 = l1;
ListNode c2 = l2;
ListNode s = new ListNode(0);
ListNode d = s;
int sum = 0;
while(c1!=null||c2!=null){
sum /= 10;
if(c1!=null){
sum += c1.val;
c1 = c1.next;
}
if(c2!=null){
sum += c2.val;
c2 = c2.next;
}
d.next = new ListNode(sum%10);
d = d.next;
}
if(sum/10==1)
d.next = new ListNode(1);
return s.next;
}
}
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
out = 0
result = ListNode(0)
l = result
while l1!=None or l2!=None:
x1 = l1.val if l1!=None else 0
x2 = l2.val if l2!=None else 0
l.next = ListNode((x1+x2+out)%10)
out = (x1+x2+out)/10
l = l.next
l1 = l1.next if l1!=None else l1
l2 = l2.next if l2!=None else l2
if out!=0:
l.next = ListNode(out)
return result.next
//better solution from discuss
class Solution(object):
def addTwoNumbers(self, l1, l2):
carry = 0
root = n = ListNode(0)
while l1 or l2 or carry:
v1 = v2 = 0
if l1:
v1 = l1.val
l1 = l1.next
if l2:
v2 = l2.val
l2 = l2.next
carry, val = divmod(v1+v2+carry, 10)
n.next = ListNode(val)
n = n.next
return root.next
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