leetcode 002

Posted 2021-01-09 今夕何夕兮

2. Add Two numbers

Difficulty:Medium

The Link

• https://leetcode.com/problems/add-two-numbers/

Description

You are given two non-empty linked lists representing two non-negative intergers. 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 -> 8
Explanation: 342 + 465 = 807.

Solutions:

Solution A:

Brute-force attack (暴力破解)
runtime: Time limit exceeded

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1:
            return l2
        if not l2:
            return l3
        val1, val2 = [l1.val],[l2.val]
        print (val1,val2)
        while l1.next:
            val1.append(l1.next.val)
        while l2.next:
            val2.append(l2.next.val)
        # [::-1] 从后往前遍历
        num1 = ‘‘.join([str(i) for i in val1[::-1]])
        num2 = ‘‘.join([str(i) for i in val2[::-1]])

        tmp = str(int(num1)+ int(num2))[::-1]
        res = ListNode(int(tmp[0]))
        run_res = res
        for i in range(1,len(tmp)):
            run_res.next = ListNode(int(tmp[i]))
            run_res = run_res.next
        return run_res

Solution B:

Recursion: runtime:32ms

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if not l1:
            return l2
        if not l2:
            return l1
        if l1.val + l2.val < 10:
            l3 = ListNode(l1.val+l2.val)
            l3.next =self. addTwoNumbers(l1.next,l2.next)
        else:
            l3 = ListNode(l1.val+ l2.val - 10 )
            tmp = ListNode(1)
            tmp.next = None
            l3.next =self. addTwoNumbers(l1.next,self.addTwoNumbers(l2.next,tmp))
       return l3