[LeetCode] 009. Palindrome Number (Easy) (C++/Java/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)

Github: https://github.com/illuz/leetcode

009.Palindrome_Number (Easy)

链接：

题目：https://oj.leetcode.com/problems/palindrome-number/

代码(github)：https://github.com/illuz/leetcode

题意：

判断一个数是否是回文数。

分析：

按自己想的去写就行了。

1. 可以先转为字符串，再判断。（这种解法用 Python 可以一句话完成 =w=）
2. 更好的方法是直接算出回文的数再直接比较。

代码：

C++:(可以先转为字符串)

class Solution {
public:
    bool isPalindrome(int x) {
		if (x < 0) return false;
		int bit[10];
		int cnt = 0;
		while (x) {
			bit[cnt++] = x % 10;
			x /= 10;
		}
		for (int i = 0; i < cnt; i++)
			if (bit[i] != bit[cnt - i - 1])
				return false;
		return true;
    }
};

Python:

class Solution:
    # @return a boolean
    def isPalindrome(self, x):
        return str(x) == str(x)[::-1]

C++:(算出回文)

class Solution {
public:
    bool isPalindrome(int x) {
		long long xx = x;
		long long new_xx = 0;

		while (xx > 0) {
			new_xx = new_xx * 10 + xx % 10;
			xx /= 10;
		}

		return new_xx == (long long)x;
    }
};

Java:

public class Solution {

    public boolean isPalindrome(int x) {
        long xx = x;
        long new_xx = 0;
        while (xx > 0) {
            new_xx = new_xx * 10 + xx % 10;
            xx /= 10;
        }
        return new_xx == x;
    }
}

Python:

class Solution:
    # @return a boolean
    def isPalindrome(self, x):
        xx = x
        new_xx = 0
        while xx > 0:
            new_xx = new_xx * 10 + xx % 10
            xx /= 10

        return new_xx == x