Topic information
source address ： Palindrome number
Give you an integer x, If x Is a palindrome integer , return true; otherwise , return false.
Palindrome number refers to positive order （ From left to right ） Reverse order （ From right to left ） Read all the same integers .
for example ,121 It's palindrome. , and 123 No .
Prompt information
Example 1
Input ：x = 121
Output ：true
Example 2
Input ：x = -121
Output ：false
explain ： Read left to right , by -121 . Read right to left , by 121- . So it's not a palindrome number .
Example 3
Input ：x = 10
Output ：false
explain ： Read right to left , by 01 . So it's not a palindrome number .
Tips
-2^31 <= x <= 2^31 - 1
Implementation logic
Double pointer
The simpler way of thinking is , First convert an integer into a string , Then use the first and last double pointers to determine whether the characters at the indicated position are the same , Until all the characters are compared .
As shown in the figure above , Using double pointers is an intuitive idea , It's important to note that , Negative integers are not palindromes , There is a difference between odd and even numbers in palindrome judgment .
package cn.fatedeity.algorithm.leetcode;
public class PalindromeNumber {
public boolean answer(int x) {
if (x < 0) {
return false;
}
// Convert to string
String s = Integer.toString(x);
// Double pointer judgment
int len = s.length();
for (int i = 0; i < len >> 1; i++) {
if (i != len - 1 - i && s.charAt(i) != s.charAt(len - 1 - i)) {
return false;
}
}
return true;
}
}
Do not convert to string
If you add the limitation that integers cannot be converted into strings to this problem , We need to think about the solution from the perspective of Mathematics .
Since you want to judge whether an integer is a palindrome , You can directly turn an integer into another integer , Then match these two integers equally , This method has more cycles than double pointers , But it takes no time to convert integers into strings , Integer efficiency is higher .
package cn.fatedeity.algorithm.leetcode;
public class PalindromeNumber {
public boolean answer(int x) {
if (x < 0) {
return false;
}
// Flip integers
int reverse = 0;
int rest = x;
while (rest >= 10) {
reverse = reverse * 10 + rest % 10;
rest = (int) (rest / 10.0);
}
return reverse * 10 + rest == x;
}
}
