Title Description

Gives a 32-bit signed integer, you need this integer number on each inverted.

Example 1:

Input: 123

Output: 321

Example 2:

Input: -123

Output: -321

Example 3:

Input: 120

Output: 21

Thinking

We can build a single digit integer reversed once. In doing so, we can check in advance whether another original integer additional digit will cause an overflow.

key point

Pop and push into digital

　　　　Pop: int X = POP 10% ;

　　　　Push: ANS ANS = + 10 * POP;

　　　　　　　x /= 10;

Check before overflow

If you do not consider this question overflow problem it is very simple. There are two ideas to solve the overflow problem, the first idea is to convert a string by adding try catch to address the way, the second idea is through mathematical calculations to solve.
Due to the lower string conversion efficiency and use more library function, so the program does not consider the problem-solving method, but to solve mathematical calculations.
By circulating the number x of every open, in calculating the new value every step of determining whether or not overflow.
Two overflow conditions, that is greater than a maximum integer NAX_VALUE, and the other is an integer less than the minimum MIN_VALE, provided the current calculation result is ANS, next to pop.

From ans * 10 + pop> MAX_VALUE view this overflow condition

　　　　When ans> MAX_VALUE / 10 and still need to add pop occurs, be sure overflow

　　　　When there ans == MAX_VALUE / 10 and pop> 7, it must overflow, 7 2 ^{31 is} bits -1

From ans * 10 + pop <MIN_VALUE view this overflow condition

　　　　When ans <MIN_VALUE / 10 and there is need to add pop occurs, some overflow

　　　　When there ans == MIN_VALUE / 10 and pop <-8, then a certain overflow, 8 -2 ^{31 is} a single-digit

Code:

class Solution {
     public  int Reverse ( int x) {
         int ANS = 0 ;
         the while (! x = 0 ) {
             // repeat the "pop" of the last digit x 
            int POP x = 10% ;
             // check before the overflow 
            if (ANS> Integer.MAX_VALUE / 10 || (ANS Integer.MAX_VALUE == / 10 && POP>. 7 ))
                 return 0 ;
             IF (ANS <of Integer.MIN_VALUE / 10 || (ANS of Integer.MIN_VALUE == / 10 && POP <-8 ))
                 return 0 ;
             //* 10-digit expansion of an equivalent, + pop is equivalent to a bit assignment 
            ANS ANS * = 10 + POP;
             // delete single digits 
            the X-/ = 10 ; 
        } 
        return ANS; 
    } 
}

LeetCode Brush title - integer reversal (simple)