# [Leetcode#7] Reverse Integer

Source: Internet
Author: User
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Reverse A integer is a very common task we need to understand for being a good programmer. It's very easy at some aspects, however, it also needs a prudent mind to understand all corner cases it may encounter.

The basic idea of reversing a number are to use mod "%" and divID "/" collectedly.
1. Get the last digit of a number (ten as base, decimal numeral system)
digit = num% 10
2. Chop off the last digit of a number
Num_remain = NUM/10

The basic idea is:

`int ret = 0 while (num_remain! = 0 ) {= num% 10= RET * + digit;= Num_remain/10;}`

However, there is a big pitfall at here. The problem is, the ret_num could exceed the max number a integer could represent in Java. What's more, if the NUM is a negative number, the symbol of the ret_num could isn't be predicted. We need to add some fix for the basic structure above.

1. Convert the number into positive form before reversion.

`if (Num < 0= num *-1;true ; }`

2. Add proper mechanism to check possible overflow. (Note:the Ret_num is positive now!)

` if  (ret! = 0) { if  (integer.max_value-digit)/ret < 10)  return  0;  if  (Neg_flag = = true   if  ( -10 < (integer.min_value + digit)/ ret)  return  0;                         }} RET  = ret * + digit; `

The reason is: (when overflow happens)
IFF Neg_flag = Fase, (note:the ret_num is positive now!)
RET * + digit > Integer.max_value <=> (integer.max_value-digit)/RET < 10
IFF Neg_flag = True,
-(RET * + digit) < Integer.min_value <=> -10 < (integer.min_value + digit)/RET

Questions:

Reverse digits of an integer.

EXAMPLE1:X = 123, return 321
example2:x = -123, return-321

`//Take care of overflow//both mod and divid can have positive/negative symbols Public classSolution {//Watch out for all corner cases!!!     Public intReverseintx) {intRET = 0; intDigit = 0; BooleanNeg_flag =false; if(X < 0) {Neg_flag=true; X=-1 * x;//Covert to ABS (x), and record the symbol of negative or positive.        }                     while(x! = 0) {digit= x% 10;//get The last digit of X                        if(ret! = 0) {//must follow this pattern to check                if((integer.max_value-digit)/RET < 10 )                     return0; if(Neg_flag = =true) {                    if( -10 < (integer.min_value + digit)/ret)//-(ret * + digit) < Integer.min_value//If we convert the number to ABS, we need to compare it in negative form with Integer.min_value                   return0; }} RET= RET * 10 +Digit; X= X/10;//chop off The last digit of X        }                if(Neg_flag = =true&& ret > 0) ret=-1 *ret; returnret; }}`

[Leetcode#7] Reverse Integer

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