README_EN.md

9. Palindrome Number

中文文档

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

Python3

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        y, t = 0, x
        while t:
            y = y * 10 + t % 10
            t //= 10
        return x == y

Java

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) return false;
        int y = 0, t = x;
        while (t != 0) {
            y = y * 10 + t % 10;
            t /= 10;
        }
        return x == y;
    }
}

JavaScript

/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function (x) {
    let str = x + '';
    let left = 0,
        right = str.length - 1;
    while (left < right) {
        if (str[left] != str[right]) return false;
        left++;
        right--;
    }
    return true;
};

Go

func isPalindrome(x int) bool {
	if x < 0 {
		return false
	}
	result := 0
	y := x
	for y != 0 {
		result = result * 10 + y%10
		y /= 10
	}
	return result == x
}

Rust

impl Solution {
    pub fn is_palindrome(x: i32) -> bool {
        if x < 0 {
            return false;
        }
        let s = x.to_string();
        let bs = s.as_bytes();
        let n = bs.len();
        let mut l = 0;
        let mut r = n - 1;
        while l < r {
            if bs[l] != bs[r] {
                return false;
            }
            l += 1;
            r -= 1;
        }
        true
    }
}

impl Solution {
    pub fn is_palindrome(mut x: i32) -> bool {
        if x < 0 || (x % 10 == 0 && x != 0) {
            return false;
        }
        let mut y = 0;
        while x > y {
            y *= 10;
            y += x % 10;
            x /= 10;
        }
        x == y || x == y / 10
    }
}

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