【原】LeetCode之Reverse Integer

1、题目

Reverse digits of an integer.

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

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Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

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2、代码实现

代码实现1、

通过不了LeetCode

    public static int reverse(int x) {
        if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {
        return 0;
        }
    boolean flag = false; //x是负数就是true，正数false
    if (x < 0) {
    flag = true;
    }
    String string = String.valueOf(x);
    String spliteStr =  flag ? string.substring(1) : string;
    StringBuffer sb = new StringBuffer(spliteStr);
    sb = sb.reverse();
    String result = sb.toString();
    result = flag ? "-" + result : result;
    long value = Long.valueOf(result);
        if (value > Integer.MAX_VALUE || value < Integer.MIN_VALUE) {
        return 0;
        }
        return (int)value;
    }

代码实现二、

通过不了LeetCode

 

    public static int reverse1(int x) {
    if (x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) {
        return 0;
        }
    long result = 0;
    int temp = Math.abs(x);
    while (temp > 0) {
    result *= 10;
    result = result + temp % 10;
    temp /= 10;
    }
    
 if (result > Integer.MAX_VALUE ||result<Integer.MIN_VALUE ) {
return 0;
}
    return (int)(x >= 0 ? result : -result);
    }

 

 

代码实现三

可以通过LeetCode

    public static int reverse3(int n) {
    if (n > Integer.MAX_VALUE || n < Integer.MIN_VALUE) {
        return 0;
        }
    //输出结果定义为long
long sum=0;
while (n != 0) {
int s = n % 10;
sum = sum * 10 + s;
n = n / 10;
}
//防止溢出操作
if (sum > Integer.MAX_VALUE || sum < Integer.MIN_VALUE) {
return 0;
}
return (int)sum;
    }

注意有溢出问题，对比分析，第一个实现和第二个实现 不越界没问题。