The seventh title integer power button reverse

Topic Overview

• Title: The seventh title
• Difficulty: Easy
• Content: Gives a 32-bit signed integer, you need this integer numbers on each inverted. Example 1:

  Input: 123 Output: 321

  Example 2:

  Input: Output -123: -321

  Example 3:

  Input: 120 Output: 21

  note:

  Suppose we have an environment can only store a 32-bit signed integer, then the value range of [-231 231--1]. Please According to this hypothesis, if integer overflow after reverse it returns 0.

  Source: stay button (LeetCode) link: https: //leetcode-cn.com/problems/reverse-integer

The first idea

The first considers only three integers, not thinking the range of integer problems.

Code

class Solution {
public:
    int reverse(int x) {
    int bai;
    int shi;
    int ge,y;
    bai = x/100;
    shi = x/10%10;
    ge = x%10;
    y=ge*100+shi*10+bai;
    if(x<0)
    {
        return y=-y;
    }
        return y;
    }  
};

Test Submit

The test results can only meet three integers, does not satisfy the other conditions

analysis

There are two points title Note:

1. 32-bit signed integer, the range from the negative 2,147,483,648 worth (represented by Int_MIN constant) to the positive 2,147,483,647 (represented by Int_MAx constant) values, not just three integer
2. Reverse way

Improve

1. Number -9 to 9, can be directly output

2. Overflow numbers, 0 is returned;

3. The intermediate portion using a while loop to solve.

   ①. After a number of defining a new long integer to store reverse

   ②. Of x modulo operation performed, then the structures are passed to y. For example: 321, modulo 10 for 1, 1 is transmitted to the y, x rounded to 10, 32; in the second cycle, the y * 10, is equal to 10, then repeat the operation, y = 12, x = 3, y = 123, x = 0, then the loop is exited

Improved Code

class Solution {
public:
    int reverse(int x) {
        long y=0;
        if(x>-9&&x<9){
            return x;
        }
        
        
        else{
            while(x){
                y=y*10;
                 if(y<=INT_MIN||y>=INT_MAX)
                return 0;//溢出
                y=y+x%10;
                x=x/10;
                
            }
            return y;
        }
    }
};

Improved Submit

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Harvest summary

The subject is relatively simple, but there are details that need to be considered, such as overflow problems, as well as the code to simplify the problem.