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：我在写完代码之后才发现最后的Follow up，真是醉了，首先想到的办法就是follow up中提到的“将整数转换为string，然后逆置一下，判断是否相等”
代码如下：
package com.Ryan;
public class PalindromeNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
PalindromeNumber palindromeNumber=new PalindromeNumber();
int x=-121;
System.out.println(palindromeNumber.isPalindrome(x));
}
public boolean isPalindrome(int x) {
String s=String.valueOf(x);
int length=s.length();
String reverse = "";
for (int i = 0; i < length; i++) {
reverse = s.charAt(i) + reverse;
}
System.out.println(reverse);
if (s.equals(reverse)) {
return true;
}
else {
return false;
}
}
}
如果不将int转换为string的话也是可以的，思路差不多，一个数学公式可以解决
int rev=0;
rev=rev*10+x%10;
x=x/10;
这样也可以实现int的逆置，这里需要在代码一开始排一下雷，例如题目中提到的负号问题，代码如下：
package com.Ryan;
public class PalindromeNumber {
public static void main(String[] args) {
// TODO Auto-generated method stub
PalindromeNumber palindromeNumber=new PalindromeNumber();
int x=-121;
System.out.println(palindromeNumber.isPalindrome(x));
}
public boolean isPalindrome(int x) {
if (x<0) {
return false;
}
int tmp=x;
int rev=0;
while (tmp!=0) {
rev=rev*10+tmp%10;
tmp=tmp/10;
}
if (rev==x) {
return true;
}
else {
return false;
}
}
}