9. Palindrome Number
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?
Solutions
Solution 1: Reverse Half of the Number
First, we determine special cases:
- If $x < 0$, then $x$ is not a palindrome, directly return
false
; - If $x > 0$ and the last digit of $x$ is $0$, then $x$ is not a palindrome, directly return
false
; - If the last digit of $x$ is not $0$, then $x$ might be a palindrome, continue the following steps.
We reverse the second half of $x$ and compare it with the first half. If they are equal, then $x$ is a palindrome, otherwise, $x$ is not a palindrome.
For example, for $x = 1221$, we can reverse the second half from "21" to "12" and compare it with the first half "12". Since they are equal, we know that $x$ is a palindrome.
Let's see how to reverse the second half.
For the number $1221$, if we perform $1221 \bmod 10$, we will get the last digit $1$. To get the second last digit, we can first remove the last digit from $1221$ by dividing by $10$, $1221 / 10 = 122$, then get the remainder of the previous result divided by $10$, $122 \bmod 10 = 2$, to get the second last digit.
If we continue this process, we will get more reversed digits.
By continuously multiplying the last digit to the variable $y$, we can get the number in reverse order.
In the code implementation, we can repeatedly "take out" the last digit of $x$ and "add" it to the end of $y$, loop until $y \ge x$. If at this time $x = y$, or $x = y / 10$, then $x$ is a palindrome.
The time complexity is $O(\log_{10}(n))$, where $n$ is $x$. For each iteration, we will divide the input by $10$, so the time complexity is $O(\log_{10}(n))$. The space complexity is $O(1)$.
1 2 3 4 5 6 7 8 9 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 |
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1 2 3 4 5 6 7 8 9 10 |
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1 2 3 4 5 6 7 8 9 10 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
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