[LeetCode]7 Reverse Integer

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

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

My accepted 15 lines of code for Java

Only 15 lines.
If overflow exists, the new result will not equal previous one.
No flags needed. No hard code like 0xf7777777 needed.
Sorry for my bad english.

public int reverse(int x){    int result = 0;    while (x != 0)    {        int tail = x % 10;        int newResult = result * 10 + tail;        if ((newResult - tail) / 10 != result)        { return 0; }        result = newResult;        x = x / 10;    }    return result;}

Very Short (7 lines) and Elegant Solution

public int reverse(int x) {        long rev= 0;        while( x != 0){            rev= rev*10 + x % 10;            x= x/10;            if( rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE)                return 0;        }        return (int) rev;    }

C版本解法：

int reverse(int x) {
    int rev = 0;
    while(x != 0) {
        int temp = rev * 10 + x % 10;
        if(temp / 10 != rev)
            return 0;
        rev = temp;
        x = x / 10;
    }
    return rev;
}