【LeetCode刷题记录】9. 回文数
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2022-03-13 12:58:47
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题目描述:
题解:
思路比较简单,引用官方题解:
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
}