7. Reverse Integer

最新推荐文章于 2025-03-06 22:12:29 发布

is.lizhichao

最新推荐文章于 2025-03-06 22:12:29 发布

阅读量150

点赞数

分类专栏： LeetCode 文章标签： leetcode

@ACHAO

本文链接：https://blog.youkuaiyun.com/sinat_41258771/article/details/105587874

版权

LeetCode 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种算法，用于翻转一个32位有符号整数的数字，包括正数和负数。提供了C++和Go语言的实现代码，确保在32位整数范围内正确处理溢出情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Reverse Integer

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

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which 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.

C++解法

class Solution {
public:
	int reverse(int x) {
		bool flag = true;
		flag = x > 0 ? true : false;
		string num = to_string(x);
		if (*num.begin() == '-') num.erase(num.begin());
		std::reverse(num.begin(), num.end());
		while (*num.begin() == '0')
			num.erase(num.begin());
		long long temp = 0;
		temp = atoll(num.c_str());
		temp = flag ? temp : -temp;
		if (temp > INT_MAX || temp < INT_MIN) return 0;
		return temp;
	}
};

Go解法

func reverse(x int) int {
    //math.MinInt32 = -2147483648
    //math.MaxInt32 = 2147483647
    var result int
    for x!=0{
        result=result*10+x%10
        if result>math.MaxInt32 || result<math.MinInt32{
            return 0
        }
        x/=10
    }
    return result
}