我的思路:
转化为字符串
对正数:
设置双指针,一个负责前半段,一个负责后半段
比较每一组对称位置是否数值相等
时间复杂度为O(N)
class Solution:
def isPalindrome(self, x: int) -> bool:
if x < 0:
return False
elif x == 0:
return True
tempstr = str(x)
#thislist = [letter for letter in tempstr]
apoint = 0
bpoint = len(tempstr) - 1#len(thislist) - 1
while apoint < bpoint:
if tempstr[apoint] != tempstr[bpoint]:
return False
else:
apoint += 1
bpoint -= 1
continue
return True