LeetCode 2. 两数相加

2. 两数相加 add-two-numbers（中）

考察知识点

链表的声明

核心思想

设立一个表示进位的变量carried，建立一个新链表，把输入的两个链表从头往后同时处理，每两个相加，将结果加上carried后的值作为一个新节点到新链表后面。

图示

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注意事项

• C++链表中的操作都是基于指针进行的。指针指向的是ListNode。

• 当链表以struct声明时，用->获取其元素。

  ListNode* p = l1;
  cout << p->val << endl;
  

代码

方法一

class ListNode():
    def __init__(self, val):
        if isinstance(val,int):
            self.val = val
            self.next = None
            
        elif isinstance(val,list):
            self.val = val[0]
            self.next = None
            cur = self
            for i in val[1:]:
                cur.next = ListNode(i)
                cur = cur.next
    
    def gatherAttrs(self):
        return ", ".join("{}: {}".format(k, getattr(self, k)) for k in self.__dict__.keys())

    def __str__(self):
            return self.__class__.__name__+" {"+"{}".format(self.gatherAttrs())+"}"


class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        if isinstance(l1,list):
            l1 = ListNode(l1)
            l2 = ListNode(l2)
        re = ListNode(0)
        r=re
        carry=0
        while(l1 or l2):
            x= l1.val if l1 else 0
            y= l2.val if l2 else 0
            s=carry+x+y
            carry=s//10
            r.next=ListNode(s%10)
            r=r.next
            if(l1!=None):l1=l1.next
            if(l2!=None):l2=l2.next
        if(carry>0):
            r.next=ListNode(1)
        return re.next

if __name__ == "__main__":
    test = Solution()
    print(test.addTwoNumbers([1,3],[2,1,3]))

• 方法二

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    # maybe standard version
    def _addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = dummy = ListNode(-1)
        carry = 0
        while l1 and l2:
            p.next = ListNode(l1.val + l2.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            l1 = l1.next
            l2 = l2.next
        
        res = l1 or l2
        while res:
            p.next = ListNode(res.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            res = res.next
        if carry:
            p.next = ListNode(1)
        return dummy.next

    # shorter version
    def addTwoNumbers(self, l1, l2):
        p = dummy = ListNode(-1)
        carry = 0
        while l1 or l2 or carry:
            val = (l1 and l1.val or 0) + (l2 and l2.val or 0) + carry
            carry = val / 10
            p.next = ListNode(val % 10)
            l1 = l1 and l1.next
            l2 = l2 and l2.next
            p = p.next
        return dummy.next


if __name__ == "__main__":
    print("hello world!")
    solution = Solution()
    l1 = ListNode([2,4,3])
    l2 =
    ListNode([5,6,4])
    solution.addTwoNumbers(l1, l2)

总结

这个题目做的太久了，主要犯了以下几个错误！

1、把for (int i = 0; i<len1-len2; i++)或for (int i = 1; i<=len1-len2; i++)错写成for (int i = 1; i<len1-len2; i++)一字之差搞了半天。