refactor: parse int as number or big int

[NguyenTuanCanh]

Development related changes

1.Use Number.isInteger to check if the input can be parsed as an integer, and if not, we parse it as a float.
2.Use Number(x) to convert the string to a number for checking integer type.
3.Split the condition and result into separate lines for improved readability.
4.Use Number.isSafeInteger to check if the parsed integer is within the safe integer range.
4.If it's a safe integer, we return it as a regular number; otherwise, we return it as a BigInt.

Checklist:

• Performed a self-review of the code
• Rebased to the last commit of the target branch (or merged it into my branch)
• Linked the issues which this PR resolves
• Documented the changes in code (API docs will be generated automatically)
• Updated the tests
• All tests are passing

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[tabaktoni]

Thank you for a nice try.

• I do not see significant gain between regex and number solution, other than using the lib method, what can be removed by doing, or creating local isInteger(str: String)

/^-?[0-9]+$/.test(x)

https://www.measurethat.net/Benchmarks/Show/28243/0/compare-number-vs-regex-test

• Regarding readability more compact method is preferred as more convenient to read.

Please provide a convenient argument for the change
If you would like to contribute please take some of the open Issues.

[penovicp]

Also didn't notice a performance change with a benchmark that uses the different approaches within the lossless-json parser override method.