10.Palindrome Number_lowest-base palindrome-优快云博客

本文链接：https://blog.youkuaiyun.com/qinzhaokun/article/details/47682209

本文介绍两种不使用额外空间判断整数是否为回文的方法。一种是将整数转化为字符串进行对比；另一种通过数学方式逐位比较最高位与最低位是否相同。讨论了负数情况及整数反转可能带来的溢出问题。

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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.

Solution: nagative and less than 10 non-nagative poistions are be specially processed.

So1: use String,more easy way.

public class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        else if (x>=0 && x<=9) {
            return true;
        }
        else{
            String a = String.valueOf(x);
            int i = 0;
            int j = a.length()-1;
            while(i<j){
                if(a.charAt(i)!=a.charAt(j)){
                    return false;
                }
                i++;
                j--;
            }
            return true;
        }
    }
}

So2: use int. each time select highest one and lowest one to judge whether they are the same. be careful the detail.

public class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        else if (x>=0 && x<=9) {
            return true;
        }
        else {
            int base = 1;
            int k = x;
            int t = 0;
            while (true) {
                t = k/10;
                if (t == 0) {
                    break;
                }
                else {
                    base *=10;
                }
                k = k/10;
            }
            
            while(x>0) {
                int low = x%10;
                int high = x/base;
                if (low != high) {
                    return false;
                }
                x = x%base;
                base = base/100;//头和尾都去掉，是除以100
                x = x/10;
            }
            return true;
            
        }
    }
}