Description
Given an integer x, return true if x is a
palindrome
, and false otherwise.
Example 1:
Input: x = 121 Output: true Explanation: 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Constraints:
-231 <= x <= 231 - 1
Follow up: Could you solve it without converting the integer to a string?
Code
to_string
轉成字串再用 two pointer 比較。
class Solution {
public:
bool isPalindrome(int x) {
string str_x = to_string(x);
int l = 0, r = str_x.length() - 1;
while(l < r) {
if(str_x[l++] != str_x[r--]) return false;
}
return true;
}
};也可以 reverse 之後再跟原本的比較。
class Solution {
public:
bool isPalindrome(int x) {
string str_x = to_string(x);
reverse(str_x.begin(), str_x.end());
return str_x == to_string(x);
}
};Math
rev 為 x 的後半部，又因為 while(rev < x)，因此 rev 可能會比 x 大十倍。
要注意 edge case：x = 10，此時倒轉過來會是 01，會有 leading zero，所以要檢查 (x % 10 == 0 && x != 0)。
class Solution {
public:
bool isPalindrome(int x) {
if(x < 0 || (x % 10 == 0 && x != 0)) return false;
int rev = 0;
while(rev < x) {
rev = rev * 10 + x % 10;
x = x / 10;
}
return x == rev || x == rev / 10;
}
};