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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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.
这道题稍微麻烦点的就是因为链表是一个逆序的,两个链表的长度可能会不一样,所以得先走到底之后把各个位存下来再相加
我就主要同时迭代两个链表,然后用字符串来保存每个位,str=listnode.val+str,每次迭代到下个节点,把val加到当前字符串的前头,迭代完之后然后转换成数字,反向,生成一个链表
你也可以存前向的节点什么的,方法挺多的
当然也可以进一个节点就加一次,然后生成一个节点,只不过这样子得处理一下进位
源码给出两种方法
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# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: node_now1=l1 node_now2=l2 str1,str2='','' while node_now1 or node_now2: if node_now1: str1=str(node_now1.val)+str1 node_now1=node_now1.next if node_now2: str2=str(node_now2.val)+str2 node_now2=node_now2.next num1=int(str1) num2=int(str2) res_int=num1+num2 # the reversion of the res_int res_str_rvrse=str(res_int)[::-1] root=ListNode(-1) cur_node=root for k in range(len(res_str_rvrse)): cur_node.val=int(res_str_rvrse[k]) if k != len(res_str_rvrse)-1: cur_node.next=ListNode(-1) cur_node=cur_node.next else: cur_node.next=None return root
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: exp,c,ans=0,0,ListNode(None) temp=ans while l1 or l2: n1=0 if not l1 else l1.val n2=0 if not l2 else l2.val n1+=n2+c c=n1//10 n1=n1%10 temp.next=ListNode(n1) exp+=1 if l1: l1=l1.next if l2: l2=l2.next temp=temp.next if c: temp.next=ListNode(c) return ans.next
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