LeetCode - Palindrome Number
Problem description
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
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Input: 121
Output: true
Example 2:
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Input: -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:
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Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Could you solve it without converting the integer to a string?
Analysis
Basic idea is to convert Integer to digits and use two pointer to compare them one by one, beats 41.68%.
Here’s the code:
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public class PalindromeNumber {
public boolean isPalindrome(int x) {
if (x < 0) {
return false;
}
if (x < 10) {
return true;
}
List<Integer> digits = new ArrayList<>();
while (x != 0) {
digits.add(x % 10);
x = x / 10;
}
int i = 0, j = digits.size() - 1;
while (i <= j){
if (digits.get(i++) != digits.get(j--)){
return false;
}
}
return true;
}
}
A more simple way is to reverse the number and check if they are equal, with some edge cases to deal with.
Here’s the code from LeetCode:
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class Solution {
public boolean isPalindrome(int x) {
if(x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}
int revertedNumber = 0;
while(x > revertedNumber) {
revertedNumber = revertedNumber * 10 + x % 10;
x /= 10;
}
return x == revertedNumber || x == revertedNumber / 10; // here to deal with the odd and even part
}
}
A tricky skill here is that we don’t need to reverse all of x, just half of the x is enough.
What to improve
- think about efficiency