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 h=x/div;//head
			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;
		  }
	}
	*/
	
}

评论

2 楼 bylijinnan 2012-08-22  

在isPalindromeInt2这个方法里面，x是不断减小的。例如12321，先判断头和尾的数字是否相等：1=1，那么x就变为232，x不断减少，避免溢出。

但isPalindromeInt这个方法，是计算反转后的数字与原数是否相等来判断是否回文。
考虑x=2147483647，一反转变成7463847412就超过了Integer.MAX_VALUE

1 楼 neyshule 2012-08-22  

这是怎么处理overflow的啊？