9. Palindrome Number (JAVA)

最新推荐文章于 2020-02-01 11:38:11 发布

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最新推荐文章于 2020-02-01 11:38:11 发布

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原文链接：http://www.cnblogs.com/qionglouyuyu/p/10757899.html

本文探讨了两种高效判断整数是否为回文的方法。第一种方法通过比较整数最左端与最右端的数字来判断；第二种方法则通过反转整数并与原数对比。文章详细解释了算法原理，并提供了代码实现。

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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?

法I：逐一比较最左端与最右端的数字

class Solution {
    public boolean isPalindrome(int x) {
        if(x<0) return false;
        
        int left; //the first bit of the inteter
        int right; //the last bit of the integer
        int divisor = 1; //divisor of each iteration
        int tmp = x;
        
        //determine the initial value of divisor
        while(x/divisor >= 10){
            divisor *= 10;
        }
        
        //check palindrome
        while(x>0){
            left = x/divisor; //divided by divisor to get the first bit
            right = x%10; //mod 10 to get the last bit
            if(left != right) return false;
            
            x = (x%divisor)/10; //mode divisor to remove the first bit, divided by 10 to remove the last bit
            divisor /= 100;
        }
        return true;
    }
}

数字问题注意：

1. 负数

2. 变换后以0开始（本题中，如10101）这种情况，除以divisor = 0，mod 10 = 最后那位。

有多少个0，就得经过多少次循环，才能使得divisor的长度和被除数相等。在长度不等的时候，没次都循环末尾需为0，才能符合Palindrome的判断。所以该算法仍然可以验证有0开头的情况。

法II：逆向思维，如果是parlindrome，那么求出的反转数应等于x

这个方法的优点：只遍历了整数长度的一半。

class Solution {
    public boolean isPalindrome(int x) {
        if(x<0 //negative number
           || (x%10 == 0 && x != 0)) //the check"x == reverseNum/10" has one exception: reverseNum is one-bit number but x isn't, this case only exist in x%10 == 0 && x != 0
            return false; 
        
        int reverseNum = 0;
        while(x > reverseNum){
            reverseNum = reverseNum * 10 + x%10;
            x /= 10;
        }
        if(x == reverseNum || x == reverseNum/10) return true; //check parlindrome of number with odd or even length
        else return false;
    }
}