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leetcode09-Palindrome Number之Java版本

我的leetcode之旅,该篇章主要完成使用Java实现算法。这是第9篇Palindrome Number

全部代码下载:Github链接:github链接,点击惊喜;写文章不易,欢迎大家采我的文章,以及给出有用的评论,当然大家也可以关注一下我的github;多谢;

1.题目简介:

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Some hints:

Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem “Reverse Integer”, you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.


2.我的思路:

1.将x的从个位开始取出乘以10+上后面的数,并判断最后是否等于X就行

3.我的AC代码

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public class Solution {
public boolean isPalindrome(int x) {
if (x < 0 || (x > 0 && x%10 == 0)) {
return false;
}
int num = 0;
while (x > num) {
num = num * 10 + x % 10;
x /= 10;
}
return x == num || num/10 == x;
}
}

好的本章介绍到这里 来自伊豚wpeace(blog.wpeace.cn)

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