題目地址:
https://www.lintcode.com/problem/palindrome-number/description
給定一個正整數,判斷其是否迴文。直接計算其翻轉後是幾,然後比較一下是否相等即可。代碼如下:
public class Solution {
/**
* @param num: a positive number
* @return: true if it's a palindrome or false
*/
public boolean isPalindrome(int num) {
// write your code here
int m = num;
int rev = 0;
while (m != 0) {
rev *= 10;
rev += m % 10;
m /= 10;
}
return num == rev;
}
}
時空複雜度。