LeetCode07, integer inversion

topic

Method: (low)

class Solution {
    
    
    public int reverse(int x) {
    
    
            if(x==0)
                return 0;
            int flag=x>0?1:-1;
            x*=flag;//转变为整数
            String x_str=Integer.toString(x);
            String reverse = new StringBuffer(x_str).reverse().toString();
            int res;

           try{
    
    
               res = Integer.parseInt(reverse);
               res = res*flag;
           }catch (NumberFormatException e){
    
    
               return 0;
           }
            return res;

    }
}

The second approach: (low)

class Solution {
    
    
     public int reverse(int x) {
    
    
          if(x==0)
            return 0;
        if(x%10==0){
    
    //不可能溢出
            if(x<0){
    
    
                x = -x;
                return -Integer.parseInt(new StringBuilder(Integer.toString(x)).reverse().toString());
            }else  return Integer.parseInt(new StringBuilder(Integer.toString(x)).reverse().toString());

        }
        int flag=x>0?1:-1;
       if(x==Integer.MIN_VALUE){
    
    
            return 0;
        }
        x = x*flag;
        //x的长度等于9

        if(x>1000000000){
    
    //有可能溢出
            String s = Integer.toString(x);
            StringBuilder rev = new StringBuilder(s).reverse();//反转跟最值比较
            StringBuilder res = new StringBuilder();
            StringBuilder sb_MIN = new StringBuilder(Integer.toString(Integer.MIN_VALUE));
            StringBuilder sb_MAX = new StringBuilder(Integer.toString(Integer.MAX_VALUE));
            if(flag<0)//负数
                res.append("-");
            res.append(rev);
            if(flag<0){
    
    
                if(res.compareTo(sb_MIN)<0)
                    return Integer.parseInt(res.toString());
            }else {
    
    
                if(res.compareTo(sb_MAX)<0){
    
    
                    return Integer.parseInt(res.toString());
                }
            }
            return 0;

        }
        return flag*Integer.parseInt(new StringBuilder(Integer.toString(x)).reverse().toString());
        }
}

The third approach: refer to the official solution. . . . . I'm too low

Take the remainder and in turn construct the inverse number of this integer. Every time a bit is constructed, it must be judged whether it overflows. The required range of the question is
assuming that our environment can only store the next 32-bit signed integer, then the value range is [−2 ³¹ , 2 ³¹ − 1]. According to this assumption, if the integer overflows after the inversion, it returns 0.
That is to say, the last bit of the marginal value is 7 or -8;

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