007.Reverse Integer

Posted on | In

Test cases

1
2
3
4
5
6
7
8
10000
123321
-2147483412
2147483412
-2147483648
2147483647
1534236469
-1534236469

Solution 1: accepted

Basicly hard code, not wise at all. 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
public class Solution {
    public int reverse(int x) {
        int reversed = 0;
    	int digit = 0;
    	while (x != 0) {
		if (digit == 9) {//2147483648
			if (reversed > 214748364 || reversed < -214748364)
				return 0;
			else if (reversed == 214748364 && x % 10 >7)
				return 0;
			else if (reversed == -214748364 && x % 10 >8)
				return 0; // not happening as input is int
		}
    		reversed = reversed * 10 + x % 10;
    		x = x / 10;
		digit++;
        }
	return reversed;
    }
}

Solution 2: accepted 41ms, 54%

Concise and fast. A smart way to detect overflow. 

1
2
3
4
5
6
7
8
9
10
11
12
public int reverse(int x) {
  int reversed = 0;
  while (x != 0) { // x could be negative
    int before = reversed;
    int tail = x % 10; // this improves speed
    reversed = reversed * 10 + tail;
    if ((reversed - tail) / 10 != before)  // check overflow
	return 0;
    x /= 10;
  }
  return reversed;
}

Solution 3: accepted 36ms 92%

Even more concise and faster! Jump out of the “box”! 

1
2
3
4
5
6
7
8
9
10
public int reverse(int x) {
  long reversed = 0;
  while (x != 0) {
    reversed = reversed * 10 + x % 10;
    if (reversed > Integer.MAX_VALUE || reversed < Integer.MIN_VALUE )
	return 0;
    x /= 10;
  }
  return (int)reversed;
}