1. Title Description

To give you one 32 Signed integer of bit x , Return to x The result of reversing the number part in .

If the integer after inversion exceeds 32 The range of signed integers of bits [−2^31 , 2^31 − 1] , Just go back to 0.
Suppose the environment doesn't allow storage 64 An integer （ With or without sign ）.

Example 1：

 Input ：x = 123
 Output ：321

Example 2：

 Input ：x = -123
 Output ：-321

Example 3：

 Input ：x = 120
 Output ：21

Example 4：

 Input ：x = 0
 Output ：0

Tips ：

-2^31 <= x <= 2^31 - 1

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/reverse-integer
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

2. Their thinking

There is nothing to say about this question , It is worth noting that int The scope of the
int The range is -2147483648-------2147483647
There are defined macros in the buckle INT_MAX and INT_MIN, But this figure seems to be used in many places , By the way .
Given is an integer x, We use a variable res Save the number after inversion
start res = 0;
Every time x A bit of p, namely p = x % 10,
Then remove the x A bit of （ Of course not , But to get x The number before the first digit , The number of digits is less than before ）, namely x /= 10;
Judge if you add p after ,res Overflow or not ,
（1） Positive numbers

 If res > INT_MAX / 10（ namely res Greater than 214748364）,
 that res multiply 10 It overflowed ,（res = res * 10 + p）, Go straight back to 0
 If res  be equal to  INT_MAX（ namely res  be equal to 214748364）,
 So judge p Is it greater than 7, If it is greater than 7,
 Then it will overflow （214748364 * 10 + 8 > 2147483647）.

（1） negative

 If res > INT_MIN / 10（ namely res Greater than -214748364）,
 that res multiply 10 It overflowed ,（res = res * 10 + p）, Go straight back to 0
 If res  be equal to  INT_MIN（ namely res  be equal to -214748364）,
 So judge p Is less than 8, If it is less than 8,
 Then it will overflow （-214748364 * 10 - 9 > -2147483649）.

3. Code

int reverse(int x) {
    
    int max = 2147483647;
    int min = -2147483648;
    
    int res = 0;
    while(x != 0){
    
        int p = x % 10;
        x /= 10;
        
        if((res > max / 10) || ((res == max / 10) && p > 7)){
    
            return 0;
        }
        else if((res < min / 10) || ((res == min / 10) && p < -8)){
    
            return 0;
        }
        else{
    
            res = res*10+p;
        }
    }
    return res;
}

result

4. Conclusion

Things have to be used more .