leetcode 力扣 两数之和

class Solution:
    def addTwoNumbers(self, l1, l2):
        n1 = []
        n2 = []
        nl = []
 
        while l1.next and l2.next:
            n1.append(l1.val)
            n2.append(l2.val)
            l1 = l1.next
            l2 = l2.next
 
        while l1.next != None:
            n1.append(l1.val)
            l1 = l1.next
        while l2.next != None:
 
            n2.append(l2.val)
            l2 = l2.next
        if l1.next==0 and l2.next==0:
            n1.append(l1.val)
            n2.append(l2.val)
        n1 = n1[::-1]
        n2 = n2[::-1]
        res1 = 0
        res2 = 0
        number1 = len(n1)
        number2 = len(n2)
        for i in n1:
            if number1 >= 0:
                res1 += i * 10 ** (len(n1) - number1)
                number1 += -1
        for i in n2:
            if number2 >= 0:
                res2 += i * 10 ** (len(n2) - number2)
                number2 += -1
 
        print(res1)
        print(res2)
 
        res3 = list(str(res1 + res2))
        res3 = res3[::-1]
        for i in res3:
            nl.append(int(i))
 
        return
思路是利用列表的特性 ,反向取值,然后计算结果 再放入列表中,倒置,
总结:revers方法可以更简洁的完成完成
题目难度较低.