C language exercise 42: Determining the number of palindromes

Determine the number of palindromes

Given an integer x, if x is a palindrome integer, return true; otherwise, return false. The palindrome number refers to the integer that can be read in the same order (from left to right) and in reverse order (from right to left).

Input: x = 121 Output: true (1)

Algorithm idea:

1. Define two variables n and tmp, initialize n to 0 to record the inverted value, initialize tmp to the original number, and use to obtain the digit;

2. Traverse tmp from low bit to high bit, and use the lowest bit, the second lowest bit... of tmp as the highest bit, the second highest bit... of n respectively.

3. Determine whether n is the same as the original number, return 1 if they are the same, otherwise return 0.

Code:

//判断回文数（ isPalindrome）
#include<stdio.h>
int isPalindrome(int x) {
	//特判负数的情况
	if (x < 0)
		return 0;
	int tmp = x;
	//定义新变量记录倒转后的值
	long long n = 0;
	while (tmp) {
		//tmp%10为当前次⾼位，作为个位数存⼊n
		n = n * 10 + tmp % 10;
		//删除个位数，前置位后移
		tmp = tmp / 10;
	}
	return x == n;
}
int main()
{
	printf("%d\n", isPalindrome(121));
	return 0;
}