Problem Description

Given a 32-bit signed integer x , return the result of inverting the digits in each bit of x.

Returns 0 if the inverted integer exceeds the range [−231, 231 − 1] for a 32-bit signed integer .

Assume the environment does not allow storing 64-bit integers (signed or unsigned).

Example 1

输入：x = 123 
输出：321

Example 2

输入：x = -123
输出：-321

Example 3

输入：x = 120
输出：21

Example 4

输入：x = 0
输出：0

problem solved

1. Ideas and Algorithms

This question is very simple if the overflow problem is not considered. There are two ways to solve the overflow problem,

The first idea is to solve it by string conversion and try catch ;
The second idea is to solve it through mathematical calculations .

Due to the low efficiency of string conversion and the use of more library functions, the solution does not consider this method, but solves it through mathematical calculations .

Each bit of the number x is disassembled by looping, and each step of calculating a new value is judged whether it overflows .
There are two overflow conditions,

One is greater than the integer maximum value MAX_VALUE ,
One is less than the integer minimum value MIN_VALUE ,

Let the current calculation result be ans , and the next bit be pop .

Judging from the overflow condition of ans * 10 + pop > MAX_VALUE

When there is ans > MAX_VALUE / 10 and there is still pop to be added, it must overflow
When ans == MAX_VALUE / 10 and pop > 7 , it must overflow, 7 is the single digit of 2^31 - 1

Judging from the overflow condition of ans * 10 + pop < MIN_VALUE

When ans < MIN_VALUE / 10 and pop needs to be added, it must overflow
When ans == MIN_VALUE / 10 and pop < -8 , it must overflow, 8 is the single digit of -2^31

2. Code`

class Solution {
    
    
    public int reverse(int x) {
    
    
        int ans = 0;
        while (x != 0) {
    
    
            int pop = x % 10;
            if (ans > Integer.MAX_VALUE / 10 || (ans == Integer.MAX_VALUE / 10 && pop > 7)) 
                return 0;
            if (ans < Integer.MIN_VALUE / 10 || (ans == Integer.MIN_VALUE / 10 && pop < -8)) 
                return 0;
            ans = ans * 10 + pop;
            x /= 10;
        }
        return ans;
    }
}

3. Graphical method

Graphical method

LeetCode simple solution - 7, integer inversion