Palindrome Number 判断一个数字是否是回文数字@LeetCode

整数回文判断算法

最新推荐文章于 2022-11-11 10:08:18 发布

原创最新推荐文章于 2022-11-11 10:08:18 发布 · 2.7k 阅读

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本文介绍了一种不使用额外空间判断整数是否为回文数的方法，通过递归比较数字的最高位与最低位来实现。文章提供了两种Java实现方案，一种利用递归和数学运算，另一种使用数组实现引用传递。

思路：

递归比较数字的最低位和最高位，向中间逼夹

package Level2;

/**
 *
 * Palindrome Number
 * 
 * Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Some hints:
Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.
 */
public class S9 {

	public static void main(String[] args) {
		System.out.println(isPalindrome(10101));
	}

	public static boolean isPalindrome(int x) {
		int digits = 0;
        int tmp = x;
        // 算出位数
        while(tmp != 0){
        	tmp /= 10;
        	digits++;
        }
        int div = (int) Math.pow(10, digits-1);
	    return rec(x, div); 
    }
	
	// 递归比较
	public static boolean rec(int x, int div) {
		if(x < 0){		// 负数
        	return false;
        }
        if(x < 10){		// 一位数
        	return true;
        }
       
        int quotient = x / div;		// 最高位
        int remainder = x % 10;	// 最低位
        if(quotient != remainder){
        	return false;
        }
        
        // 更新除数和被除数
        return rec((x-quotient*div-remainder)/10, div/100); 
	}
	
}

还是用递归，用数组来实现java中的引用传递，在x里从左往右比较，在y里从右往左比较

12321

x-> <-y

public class Solution {
    public boolean isPalindrome(int x) {
        if(x < 0) {
            return false;
        }
        int[] y = new int[1];
        y[0] = x;
        return recCompare(x,y);
    }
    
    public boolean recCompare(int x, int[] y) {
        if(x>=0 && x<=9) {
            return true;
        }
        
        if(!recCompare(x/10, y)) {
            return false;
        }
        
        if((x/10)%10 != y[0]%10) {
            return false;
        }
        
        y[0] /= 10;
        return true;
    }
}