题目链接
tag:
- easy
question：
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121
Output: true
Example 2:
Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Coud you solve it without converting the integer to a string?
思路：
这道验证回文数字的题如果将数字转为字符串，就变成了验证回文字符串的题，没啥难度了，我们就直接来做follow up吧，不能转为字符串，而是直接对整数进行操作，我们可以利用取整和取余来获得我们想要的数字，比如 1221 这个数字，如果 计算 1221 / 1000， 则可得首位1， 如果 1221 % 10， 则可得到末尾1，进行比较，然后把中间的22取出继续比较。代码如下：
C++ 解法：
class Solution {
public:
bool isPalindrome(int x) {
/*
string s = to_string(x);
int left=0, right=s.size()-1;
while(left <= right) {
if(s[left] != s[right])
return false;
++left;
--right;
}
return true;
*/
if(x < 0)
return false;
int div = 1;
while(x/div >= 10)
div *= 10;
while(x > 0) {
int left = x / div;
int right = x % 10;
if(left != right)
return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
};
Python 解法：
class Solution:
def isPalindrome(self, x: int) -> bool:
if x < 0:
return False
div = 1
while x / div >= 10:
div *= 10
while x > 0:
left = x // div
right = x % 10
if left != right:
return False
x = (x % div) // 10
div /= 100
return True
