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LeetCode——Palindrome Number

时间:2014-07-05 10:44:54      阅读:368      评论:0      收藏:0      [点我收藏+]

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

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.

判断一个整数是否是回文数。不要是使用而外的空间。

首先想到的是如下的方法,即若是个位数,则返回true,否则就前后对应的位置判断是否相等。

	public static boolean isPalindrome(int x) {
		if(x / 10 == 0 && x >= 0)
			return true;
		String str = x + "";
		char ch[] = str.toCharArray();
		for(int i=0;i<ch.length / 2;i++){
			if(ch[i] != ch[ch.length - i - 1])
				return false;
		}
		return true;
	}

也可以将原来的数逆一下,再与原来的数进行比较。

	public static boolean isPalindrome(int x) {
		int reverse = 0,temp = Math.abs(x);//考虑了负数,转成正数处理 ?
		while( x != 0 )
		   {
		      reverse = reverse * 10;
		      reverse = reverse + x%10;
		      x = x/10;
		   }
		return reverse == temp;
	}


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LeetCode——Palindrome Number

原文:http://blog.csdn.net/laozhaokun/article/details/36642693

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