leetcode|【python3】leetcode 445. Add Two Numbers II (Medium) leetcode|python

445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
【leetcode|【python3】leetcode 445. Add Two Numbers II (Medium)】You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7

1 题目要求不reverse ，所以转换成str求和，然后再转回ListNodetime O(n),space O(n)

在discussion里看到一个回复说的隐患：You cannot do this way, because, integer overflow occurs, if linked list is big enough. The main objective of linked list addition is to use for big integers.如果两个链表长一点，转成int就会超出int的范围。我觉得说的很有道理，所以虽然这种解法Accepted但是还不不太可取。

# Definition for singly-linked list. # class ListNode(object): #def __init__(self, x): #self.val = x #self.next = Noneclass Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ a = ''; b = '' while(l1): a += str(l1.val) l1 = l1.next while(l2): b += str(l2.val) l2 = l2.next c = str(int(a) + int(b)) l3 = ListNode(int(c[0])) node = l3 for i in range(1,len(c)): node.next = ListNode(int(c[i])) node = node.next return l3

2 虽然不能转成int 但还是可以先把val提出到容器中（如str或list）然后按位相加

class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ a = []; b = [] while(l1): a.append(l1.val) l1 = l1.next while(l2): b.append(l2.val) l2 = l2.next tag = 0 c = [] maxlen = max(len(a),len(b)) #对齐补0 if len(a) < maxlen: for i in range(maxlen - len(a)):a.insert(0,0) if len(b) < maxlen: for i in range(maxlen - len(b)):b.insert(0,0)for i in range(0,maxlen): sum = a[maxlen-i-1] + b[maxlen-i-1] + tag tag = 1 if sum >= 10 else 0 c.insert(0,str(sum%10)) if tag == 1: c.insert(0,1) l3 = ListNode(c[0]) node = l3 for i in range(1,len(c)): node.next = ListNode(int(c[i])) node = node.next node.next = None return l3