NumberFormatException 对于输入字符串：“9646324351”

Question

Here's a snapshot of a problem on LeetCode :

For this problem, they provide this framework code you have to fill in:

public int reverse(int x) {
}

When testing solutions, it seems to provide an unreasonable input and won't let me submit my solution. The error is:

Note that the input is 1534236469 . That fits in an int as input , but the reversed version, 9646324351 , naturally doesn't, as the max positive value of an int in Java is 2147483647 .

Is this just an error in the LeetCode test? Or is there some trick I can't imagine that magically lets me return 9646324351 , a clearly out-of-range value, back from reverse as an int ?

Here's my code, but the code almost doesn't matter, as the return type ( int ) is fixed by the problem (so by "integer" they really mean int , not long ):

public int reverse(int x) {
    String intString;
    StringBuilder sb = new StringBuilder();
    if(x < 0){
        sb.append("-");
        x = x * -1;
    }
    intString = Integer.toString(x);
    for(int i = intString.length() - 1; i >= 0; i--){
        sb.append(intString.charAt(i));
    }
    String resultString = sb.toString();
    int result = Integer.parseInt(resultString);
    return result;
}

Answer 1

This is an error in the LeetCode test. There is simply no way to store the return value they're asking for in an int , thus no way to return it from reverse per the framework code provided.

Perhaps the test cases were randomly generated across the full range of int , not taking into account that reverse can take the value out of range for an int for certain inputs.

It turns out that they want you to return 0 if the reversed number would be out of range. Not that they bother to mention that anywhere in the problem.

Answer 2

I finished by follow method

    int result=0;
    try{
        result = Integer.parseInt(stringBuilder.toString());
        return result;
    }catch(Exception e){
        return 0;
    }

Answer 3

You should use Long.parseLong() . The number is too big for integer. The maximum value for integer is 2^31-1=2147483647.

Answer 4

I have successfully submit the code but still I want to optimize in terms of running speed and space utilization. Can you please share your thoughts on this.

class Solution {
    public int reverse(int x) {
        String reverse;
        if(x==0 && x > Integer.MAX_VALUE && x < Integer.MIN_VALUE){return 0;}
        if(x<0){
            x = Math.abs(x);
            String temp = String.valueOf(x);
            int len = temp.length();
            reverse = "-";
            for(int i=len-1;i>=0;i--){
                reverse = reverse + temp.charAt(i);
            } 
        }else{
            String temp = String.valueOf(x);
            int len = temp.length();
            reverse = "";
            for(int i=len-1;i>=0;i--){
                reverse = reverse + temp.charAt(i);
            } 
        } 
        try{
        int rev = Integer.parseInt(reverse);
        return rev;
        }catch(Exception e){return 0;}
    }
}

Runtime: 10 ms, faster than 8.10% of Java online submissions for Reverse Integer. Memory Usage: 37 MB, less than 13.88% of Java online submissions for Reverse Integer.