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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

$begingroup$

Optimized from my 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) i < INT_MIN) 
 return 0;
 
 int sign = 1;
 if(i < 0) 
 sign = -1;
 i = i*sign;
 

 int reversed = 0;
 int pop = 0;

 while(i > 0) 
 pop = i % 10;
 reversed = reversed*10 + pop;
 i /= 10;
 

 std::cout << reversed << 'n';

 return reversed*sign;
 
;

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

share

asked 20 secs ago

greggreg

38218

$endgroup$

add a comment | 

0

$begingroup$

Optimized from my 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) i < INT_MIN) 
 return 0;
 
 int sign = 1;
 if(i < 0) 
 sign = -1;
 i = i*sign;
 

 int reversed = 0;
 int pop = 0;

 while(i > 0) 
 pop = i % 10;
 reversed = reversed*10 + pop;
 i /= 10;
 

 std::cout << reversed << 'n';

 return reversed*sign;
 
;

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

share

asked 20 secs ago

greggreg

38218

$endgroup$

add a comment | 

$begingroup$

Optimized from my 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) i < INT_MIN) 
 return 0;
 
 int sign = 1;
 if(i < 0) 
 sign = -1;
 i = i*sign;
 

 int reversed = 0;
 int pop = 0;

 while(i > 0) 
 pop = i % 10;
 reversed = reversed*10 + pop;
 i /= 10;
 

 std::cout << reversed << 'n';

 return reversed*sign;
 
;

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

share

asked 20 secs ago

greggreg

38218

$endgroup$

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

class Solution 

 public:
 int reverse(int i) i < INT_MIN) 
 return 0;
 
 int sign = 1;
 if(i < 0) 
 sign = -1;
 i = i*sign;
 

 int reversed = 0;
 int pop = 0;

 while(i > 0) 
 pop = i % 10;
 reversed = reversed*10 + pop;
 i /= 10;
 

 std::cout << reversed << 'n';

 return reversed*sign;
 
;

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);

share

asked 20 secs ago

greggreg

38218

share

asked 20 secs ago

greggreg

38218

asked 20 secs ago

greggreg

38218

asked 20 secs ago

greggreg

38218