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T140.java
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37 lines (33 loc) · 1018 Bytes
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/**
* @Author:Aliyang
* @Data: Created in 下午1:00 18-6-18
* palindrome-number：我的解法
* 思路：注意这里要用long，不然会溢出。先用hight不断地除x，直到为0，知道了位数，然后从高位往低位，从低位往高位不断取数比较，负数不是回文数
**/
public class T140 {
public boolean isPalindrome(int x) {
if (x<0||x!=0&&x%10==0)//负数或者10的整数倍
return false;
int y=x;
int count=0;
long hight=1,low=1;
while (y!=0){
y/=10;
count++;
hight*=10;
}
for (int i=0;i<count/2;i++){
long a=x%hight/(hight/10);//高位的数
long b=x/low%10;//低位的数
if (a!=b)
return false;
hight/=10;
low*=10;
}
return true;
}
public static void main(String[] args){
T140 t=new T140();
System.out.println(t.isPalindrome(1000000001));
}
}