[C + +] range control issues with inverted 32-bit int integers

Title Source: Leetcode Reverse Integer

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

Example 1:example 2:example 3:

input : 123 input : -123 Input: 120

output: 321 output : -321 output: 21

Note:
Assume we is dealing with a environment which could only store integers within the 32-bit signed integer range: [? 231, 231? 1]. For the purpose of this problem, assume a your function returns 0 when the reversed integer overflows.

Ideas:

• Consider three different scenarios:

1. 1. The value itself overflows.
   2. The value is going to overflow, the demarcation is the same as the int_max digit in unknown number.
   3. The value is within a safe range.

• Solution:

1. 1. The overflow is determined first, if the overflow returns directly 0
   2. Edge-by-bit comparison, side reversal, judgment overflow, if overflow returns 0
   3. Direct reverse

Prerequisite Processing:

1 int m=x,i=1, ten=1,//i record digit, ten record maximum number of 2while (m/=10 ) {++i;ten*=;} 3 if (i<) m=reversemin (x,i,ten); 4 Else M=reversemax (x,i,ten);

To determine the value overflow:

1 unsigned int y = x;//Extended Range
2 if (x<0) y=-x;
3 if (Y>int_max) return 0;

Bitwise comparison, and reverse:

1 intReversemax (intXintIintTen)2 {3     intsign=1;4     if(x<0) {x=-x; sign=-1;} Negative number becomes positive, sign restores negative number5     intm=0, max=Int_max;6      while(i--) {7             intRest = x%Ten;8             intmrest=max/Ten;9             if(rest>mrest)return 0;//compared to Int_max bitwise, if greater than is already overflow, return 0Ten             Else if(rest<mrest) Break;//If it is less than the safe range, jump to a safe reverse OneM + = rest *ten;//The next comparison is required only in equal circumstances AX/=Ten; -max=max-mrest*Ten; -Ten/=Ten; the         } -     if(ten==0)returnm*Sign ; -++i;//last loop abort, no action, so restore the number of deleted bits -      +      while(i--) {//Normal reverse -             intRest = x%Ten; +M + = rest *Ten; AX/=Ten; atTen/=Ten; -         } -     returnm*Sign ; -      -}

！ Negative number after modulo is negative, can not be compared, so first take positive.

！ Determine whether to reverse the end, the 16th line, can not use the number of digits, because the break out of the number of bits also minus 1

Direct reverse:

 1 intReversemin (intXintIintTen)2     { 3         intm =0; 4          while(i--) { 5             intRest = x%Ten; 6M + = rest *Ten;7X/=Ten; 8Ten/=Ten; 9         }Ten         returnm; One}

Overall code:

1 classSolution {2  Public:3     intReverseintx) {4Unsignedinty =x;5         if(x<0) y=-x;6        if(Y>int_max)return 0;7        intReversemin (intXintIintten);8        intReversemax (intXintIintten);9        intM=x,i=1, ten=1;Ten          while(m/=Ten) {++i;ten*=Ten;} One         if(i<Ten) m=reversemin (x,i,ten); A         Elsem=Reversemax (x,i,ten); -         returnm; -     } the }; - intReversemin (intXintIintTen) -     { -         intm =0; +          while(i--) { -             intRest = x%Ten; +M + = rest *Ten; AX/=Ten; atTen/=Ten; -         } -         returnm; - } - intReversemax (intXintIintTen) - { in     intsign=1; -     if(x<0) {x=-x; sign=-1;} to     intm=0, max=Int_max; +      while(i--) { -             intRest = x%Ten; the             intmrest=max/Ten; *             if(rest>mrest)return 0; $             Else if(rest<mrest) Break;Panax NotoginsengM + = rest *Ten; -X/=Ten; themax=max-mrest*Ten; +Ten/=Ten; A         } the     if(ten==0)returnm*Sign ; +++i; -      $      while(i--) { $             intRest = x%Ten; -M + = rest *Ten; -X/=Ten; theTen/=Ten; -         }Wuyi     returnm*Sign ; the      -}

[C + +] range control issues with inverted 32-bit int integers