Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ OptimizedReverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$

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Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ Optimized


Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$













0












$begingroup$


LeetCode Problem



Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.



Feedback



Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.



#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>

class Solution 

 public:
 int reverse(int i) 

 if(i > INT_MAX 
;

int main()

 Solution s; 

 assert(s.reverse(1) == 1);
 assert(s.reverse(0) == 0);
 assert(s.reverse(123) == 321);
 assert(s.reverse(120) == 21);
 assert(s.reverse(-123) == -321);
 assert(s.reverse(1207) == 7021); 
 assert(s.reverse(1534236469) == 0);
 assert(s.reverse(-2147483412) == -2143847412);






c++ mathematics integer integer-overflow






share











$endgroup$
















    0












    $begingroup$


    LeetCode Problem



    Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.



    Feedback



    Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.



    #include <cassert>
    #include <climits>
    #include <cmath>
    #include <iostream>

    class Solution 

     public:
     int reverse(int i) 

     if(i > INT_MAX 
    ;

    int main()

     Solution s; 

     assert(s.reverse(1) == 1);
     assert(s.reverse(0) == 0);
     assert(s.reverse(123) == 321);
     assert(s.reverse(120) == 21);
     assert(s.reverse(-123) == -321);
     assert(s.reverse(1207) == 7021); 
     assert(s.reverse(1534236469) == 0);
     assert(s.reverse(-2147483412) == -2143847412);






    c++ mathematics integer integer-overflow






    share











    $endgroup$














      0












      0








      0





      $begingroup$


      LeetCode Problem



      Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.



      Feedback



      Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.



      #include <cassert>
      #include <climits>
      #include <cmath>
      #include <iostream>

      class Solution 

       public:
       int reverse(int i) 

       if(i > INT_MAX 
      ;

      int main()

       Solution s; 

       assert(s.reverse(1) == 1);
       assert(s.reverse(0) == 0);
       assert(s.reverse(123) == 321);
       assert(s.reverse(120) == 21);
       assert(s.reverse(-123) == -321);
       assert(s.reverse(1207) == 7021); 
       assert(s.reverse(1534236469) == 0);
       assert(s.reverse(-2147483412) == -2143847412);






      c++ mathematics integer integer-overflow






      share











      $endgroup$




      LeetCode Problem



      Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.



      Feedback



      Optimized from beta code in the original question here. Based on this LeetCode problem.. I'm trying to come up with a good approach that just math operations to iterate a serious of digits using division, almost treating the int like a stack poping using modulus, and pushing using multiplication. I think there still may be some overflow edge cases I am not catching like 1534236469, but need some help coming up with how to detect and prevent such cases.



      #include <cassert>
      #include <climits>
      #include <cmath>
      #include <iostream>

      class Solution 

       public:
       int reverse(int i) 

       if(i > INT_MAX 
      ;

      int main()

       Solution s; 

       assert(s.reverse(1) == 1);
       assert(s.reverse(0) == 0);
       assert(s.reverse(123) == 321);
       assert(s.reverse(120) == 21);
       assert(s.reverse(-123) == -321);
       assert(s.reverse(1207) == 7021); 
       assert(s.reverse(1534236469) == 0);
       assert(s.reverse(-2147483412) == -2143847412);






      c++ mathematics integer integer-overflow




      c++ mathematics integer integer-overflow





      share














      share












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      edited 39 secs ago







      greg

















      asked 5 mins ago









      greggreg

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