Java coding problem (Boolean algebra and truth tables)

Write a program to generate a truth table for a given Boolean statement. For example, A + B * C produces the following table:

A B C | B * C | A + B * C

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0 0 0 | 0 | 0

0 0 1 | 0 | 0

0 1 0 | 0 | 0

0 1 1 | 1 | 1

1 0 0 | 0 | 1

1 0 1 | 0 | 1

1 1 0 | 0 | 1

1 1 1 | 1 | 1

Input from the keyboard a string representing a Boolean expression where the operators are the following: ! (NOT), * (AND), + (OR), ^ (XOR). Single sets of parentheses can be used if needed. Covert all letters to uppercase and have one space between each operator. The truth table must be able to handle up to four variables and be in binary order. For example, in the first test case, if the Boolean values of A, B, and C were replaced with 1’s for true and 0’s for false, the Boolean combination with the smallest binary representation would come first.

Output to the screen a truth table for the given Boolean expression. The first column of the truth table should represent A, the second B, the third C, and so on. Each binary operation must have its own column and all columns must be labeled. The entries in each column should be aligned and formatted with capital letters like the examples below. Except for the final column, columns do not have to be in any order except the order of operations must be followed to get the correct final answer. Finally, the program should ask if the user wants to run the program again (Check case). Refer to the sample output below.

Sample Run:

Enter the Boolean expression: A + B * C

A B C | B * C | A + B * C

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0 0 0 | 0 | 0

0 0 1 | 0 | 0

0 1 0 | 0 | 0

0 1 1 | 1 | 1

1 0 0 | 0 | 1

1 0 1 | 0 | 1

1 1 0 | 0 | 1

1 1 1 | 1 | 1

Run again (Y/N)? y

Enter the Boolean expression: (!A * B) ^ (B + C)

A B C | !A * B | B + C | (!A * B) ^ (B + C)

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0 0 0 | 0 | 0 | 0

0 0 1 | 0 | 1 | 1

0 1 0 | 1 | 1 | 0

0 1 1 | 1 | 1 | 0

1 0 0 | 0 | 0 | 0

1 0 1 | 0 | 1 | 1

1 1 0 | 0 | 1 | 1

1 1 1 | 0 | 1 | 1

Run again (Y/N)? N