leetcode中国官方网站-题库
https://leetcode-cn.com/problemset/all/
leetcode官网-题库
https://leetcode.com/problemset/all/
参考资料
https://blog.csdn.net/minione_2016/article/details/78855124
0001.Two_Sum
答案
class Solution(object):
def twoSum(self,nums,target):
dic = {}
for i, num in enumerate(nums):
if target - num in dic:
return [dic[target - num],i]
else:
dic[num] = i
0007.Reverse_Integer
答案1
class Solution(object):
def reverseInteger(self,n):
rev=0
n1 = n if n > 0 else -n
#除数余数结合
while n1:
rev = rev*10 + n1%10
n1 = n1//10
if rev > 2**31-1 or rev < -2**31:
return 0
return rev if n >0 else -rev
答案2
class Solution(object):
def reverseInteger(self,n):
n1 = n if n > 0 else -n
#思路：（1）int转为str;(2）str转为list;(3)翻转list;(4)list转str;(5)str转int
str1 = str(n1)
list1 = list(str1)
list2 = list1[::-1]
str2 = ''.join(list2)
n2 = int(str2)
if n2 > 2**31-1 or n2 < -2**31:
return 0
return n2 if n > 0 else -n2
0009.Palindrome_Number
答案1
class Solution(object):
def reverseInteger(self,n):
#思路：（1）int转为str;(2）str转为list;(3)翻转list;(4)判断翻转前后的list是否相同
str1 = str(n)
list1 = list(str1)
list2 = list1[::-1]
return True if list2 == list1 else False
答案2（进阶）
class Solution(object):
def reverseInteger(self,n):
#负数一定不是回文数
if n < 0:
return False
#思路：除数余数结合，根据翻转后的值与原值是否相等判断该值是否为回文数
rev = 0
n1 = n
while n1 > 0:
rev = rev*10 + n1%10
n1 = n1//10
return rev == n