7. Reverse Integer.md

7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2^{31 − 1}]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

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Sulution

• mine
  • Java

    • String Runtime: 2 ms, faster than 23.53%, Memory Usage: 33.7 MB, less than 11.66%

      public int reverse(int x) {
          boolean negative = x < 0;
          x = Math.abs(x);
          String str = String.valueOf(x);
          StringBuilder sb = new StringBuilder();
          for (int i = str.length() - 1; i >= 0; i--) {
              sb.append(str.charAt(i));
          }
          try {
              int res = Integer.parseInt(sb.toString());
              return negative ? 0 - res : res;
          } catch (Exception e) {
              return 0;
          }
      }
      

    • Math Runtime: 1 ms, faster than 100.00%, Memory Usage: 33.8 MB, less than 11.11%

      public int reverse(int x) {
          if (x == Integer.MAX_VALUE || x == Integer.MIN_VALUE){
              return 0;
          }
          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;
      }
      

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• the most votes

  • Math Runtime: 1 ms, faster than 100.00%, Memory Usage: 33.5 MB, less than 11.66%

    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;
    }
    

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