題目描述:
題解:
思路比較簡單,引用官方題解:
bool isPalindrome(int x) {
if (x < 0 || (x % 10 == 0 && x != 0)) return false;
int reverted_num = 0;
while (x > reverted_num) {
reverted_num = reverted_num * 10 + x % 10;
x /= 10;
}
return x == reverted_num || x == reverted_num / 10; //even/odd
}