[LeetCode] Palindrome Number

最新推荐文章于 2020-08-18 15:23:01 发布

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本文介绍了一种不使用额外空间判断整数是否为回文的方法。通过计算整数位数并对比首位数字来验证其回文特性，同时考虑了边界情况如负数和整数溢出问题。

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.

public class Palindrome {
	
	public static void main(String[] args) {
		System.out.println(isPalindrome(1000000001));
	}
	
	public static boolean isPalindrome(int x){
		
		//防止越界
		if (x > Integer.MAX_VALUE || x < 0) {
			return false;
		}
		//0的时候是回文
		if (x == 0) {
			return true;
		}
		
		//1.计算位数
		int temp  = x;
		int icount = 0;
		while(temp > 0){
			temp = temp / 10;
			icount++;
		}
		
		boolean isPalindrome = true;
		//2.比较
		for(int i = 0; i < icount; i++){
			//拿到对应的左位和右位
			int left = x / (int)Math.pow(10, icount - i - 1) % 10;
			int right = x / (int)Math.pow(10, i) % 10;
			if (left != right) {
				isPalindrome = false;
			}
		}
		
		return isPalindrome;
	}
}