Leikou question 9-number of palindrome

This problem can make use of question 7 - digit reversal

Then compare the reversed number with the original number. If they are equal, it is a palindrome.

Solution 1: Use the previous number inversion

class Solution {
    
    
    public boolean isPalindrome(int x) {
    
    
        if(x<0){
    
    
            return false;
        }
        int t=x;
        int ans = 0;
        while (x != 0) {
    
    //说明没有移除完
            if ((ans * 10) / 10 != ans) {
    
     // 处理过程中如果遇到越界则会返回0；
                ans = 0;
                break;
            }
            ans = ans * 10 + x % 10; // 得到反转的整数值
            x = x / 10;//移除一位
        }
        return t == ans;
    }
}

Solution 2: For example, java api (recommended in this way)

Use the API provided by StringBuilder and StringBuffer

class Solution {
    
    
    public boolean isPalindrome(int x) {
    
    
        if(x<0){
    
    
            return false;
        }
        String reverseStr = new StringBuilder(String.valueOf(x)).reverse().toString();
        return reverseStr.equals(String.valueOf(x));
    }
}

Solution two is simple, but requires a certain degree of knowledge. First, you must be familiar with StringBuilder and StringBuffer, and know that there is a reverse method