7. Reverse Integer

LeetCode
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My Solution (StringBuilder)

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class Solution {
    public int reverse(int x) {
        String str = String.valueOf(x);
        StringBuilder sb = new StringBuilder();
        int end = 0;
        if(str.charAt(0) == '-') {
            end++;
            sb.append('-');
        }
        for(int pos = str.length() - 1; pos >= end; pos--){
            sb.append(str.charAt(pos));
        }
        try {
            int number = Integer.parseInt(sb.toString());
            return number;
        } catch (NumberFormatException e) {
            return 0;
        }


    }
}

Remarks:

  1. Be careful of int boundary; use try and catch to hold big numbers error.
  2. Java get char: str.valueOf(num);
  3. Java convert to string: String.valueOf(x)
  4. Java convert to int: Integer.parseInt(str)
  5. TC: $O(n)$

Math

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class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (
                rev > Integer.MAX_VALUE / 10 ||
                (rev == Integer.MAX_VALUE / 10 && pop > 7)
            ) return 0;
            if (
                rev < Integer.MIN_VALUE / 10 ||
                (rev == Integer.MIN_VALUE / 10 && pop < -8)
            ) return 0;
            rev = rev * 10 + pop;
        }
        return rev;
    }
}

Remarks:

  1. Be careful of int boundary; check before x10: compare rev and Integer.MAX_VALUE/10.
  2. TC: $O(\log(x))$