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【LeetCode题解】Reverse Integer

2019年08月31日 ⁄ 综合 ⁄ 共 970字 ⁄ 字号 评论关闭

Reverse Integer(Java代码)

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

解题思路:我是把整数变成字符串,先判断正负,再进行反转,然后去掉开头的“0”,如果是负数再加上符号。

public class Solution {
	public int reverse(int x) {
		String str = String.valueOf(x);
		if (str.startsWith("-")) {
			str = str.substring(1, str.length());
			str = new StringBuffer(str).reverse().toString();
			str = str.replace("^0+", "");

			x = Integer.parseInt(str);
			x = -x;
		} else {
			str = new StringBuffer(str).reverse().toString();
			str = str.replace("^0+", "");
			x = Integer.parseInt(str);
		}

		return x;

	}
}

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