leetcode-7-Reverse-Integer

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描述

Reverse digits of an integer.

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

分析

把一个数字倒过来，题目不难，需要考虑的是边界情况，如果越界，返回0

解决方案1（python）

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class Solution:
    def reverse(self, x: int) -> int:
        INT_MIN, INT_MAX = -2**31, 2**31-1
        
        result = 0
        while x != 0:
            if result < INT_MIN // 10 + 1 or result > INT_MAX // 10:
                return 0
            nowDigit = x % 10
            if x < 0 and nowDigit < 10:
                nowDigit -= 10
            x = (x - nowDigit) // 10
            result = result * 10 + nowDigit
        return result

解决方案2(C++)

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class Solution {
public:
    int reverse(int x) {
        long long result = 0;
        
        while(x != 0) {
            int now = x % 10;
            result = result * 10 + now;
            if(result > INT_MAX || result < INT_MIN) {
                return 0;
            }
            x = x / 10;
        }
        return result;
    }
};

解决方案3(Rust)

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impl Solution {
    pub fn reverse(x: i32) -> i32 {
        x.signum() * x
            .abs()
            .to_string()
            .chars()
            .rev()
            .collect::<String>()
            .parse::<i32>()
            .unwrap_or(0)
    }
}

解决方案4（Golang）

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

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