LeetCode - Easy - 7. Reverse Integer

Topic

• Math

Description

https://leetcode.com/problems/reverse-integer/

Given a 32-bit signed integer, reverse digits of an integer.

Note:

Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹ − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Example 4:

Input: x = 0
Output: 0

Constraints:

• -2³¹ <= x <= 2³¹ - 1

Analysis

经验教训：

判断溢出时的老路：

boolean sign = x < 0;
...
if(sign && result > 0 || !sign && result < 0) {
	return 0;
}

但是用在本题就不好用了。

System.out.println(Integer.MAX_VALUE + 1);// -2147483648
System.out.println(964632435 * 10);// 1056389758

964632435 * 10明明溢出了，却还是正数。猜测原因是964632435 * 10底层算法为10个964632435进行累加，验证代码如下：

int result = 0;
for (int i = 0; i < 10; i++)
	result += 964632435;
System.out.println(result);// 1056389758

猜测正确。

Submission

public class ReverseInteger {
	public int reverse(int x) {
		int result = 0, last = 0;
		while (x != 0) {
			last = result;
			result = result * 10 + x % 10;
			if (result / 10 != last) {
				return 0;
			}
			x /= 10;
		}
		return result;
	}

	public static void main(String[] args) {
		System.out.println(Integer.MAX_VALUE + 1);// -2147483648
		System.out.println(964632435 * 10);// 1056389758
		int result = 0;
		for (int i = 0; i < 10; i++)
			result += 964632435;
		System.out.println(result);// 1056389758
	}
}

Test

import static org.junit.Assert.*;
import org.junit.Test;

public class ReverseIntegerTest {

	@Test
	public void test() {
		ReverseInteger obj = new ReverseInteger();

		assertEquals(321, obj.reverse(123));
		assertEquals(-321, obj.reverse(-123));
		assertEquals(21, obj.reverse(120));
		assertEquals(0, obj.reverse(0));
		assertEquals(0, obj.reverse(1534236469));
	}
}

本文地址：https://blog.csdn.net/u011863024/article/details/112007198