# java-判断一个整数是否回文，考虑溢出

```
public class PalindromeInt {

/**
* PalindromeInt,like 1,121,12321....
* you should consider the possibility that the reversed number might overflow
* eg. 1...................9,after reversing,9 comes to first,and...you know that.
*
*/
public static void main(String[] args) {
int[] a={12321,17770};
PalindromeInt pi=new PalindromeInt();
boolean re=pi.isPalindromeInt2(a[0]);
System.out.println(re);
re=pi.isPalindromeInt2(a[1]);
System.out.println(re);
}

//generic solution
public boolean isPalindromeInt(int x){
if(x<0)return false;
int x2=x;
int y=0;
while(x>0){
y*=10;
y+=x%10;
x/=10;
}
return x2==y;
}

//better solution
//avoid overflow
public boolean isPalindromeInt2(int x){
if(x<0)return false;
boolean re=true;
int div=1;
while(x/div>=10){
div*=10;
}
while(x>0){
int t=x%10;//tail
if(h!=t){
re=false;
break;
}
x=(x%div)/10;//now x is 232 instead of 12321
div/=100;//accordingly,div should be 100 instead of 10000
}
return re;
}

/* c/c++? I don't know how it works
* invoke like that: isPalindrome(x, x)
boolean isPalindrome(int x,int &y){
if (x < 0) return false;
if (x == 0) return true;
if (isPalindrome(x/10, y) && (x%10 == y%10)) {
y /= 10;
return true;
} else {
return false;
}
}
*/

}

```

java-判断一个整数是否回文，考虑溢出

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