2.2 Ternary Expression Parser 439

Description

Given a string representing arbitrarily nested ternary expressions, calculate the result of the expression. You can always assume that the given expression is valid and only consists of digits0-9,?,:,TandF(TandFrepresent True and False respectively).

Note:

1. The length of the given string is ≤ 10000.
2. Each number will contain only one digit.
3. The conditional expressions group right-to-left (as usual in most languages).
4. The condition will always be either T or F . That is, the condition will never be a digit.
5. The result of the expression will always evaluate to either a digit 0-9 , T or F .

Example 1:

Input:
 "T?2:3"


Output:
 "2"


Explanation:
 If true, then result is 2; otherwise result is 3.

Example 2:

Input:
 "F?1:T?4:5"


Output:
 "4"


Explanation:
 The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(F ? 1 : (T ? 4 : 5))"                   "(F ? 1 : (T ? 4 : 5))"
          -
>
 "(F ? 1 : 4)"                 or       -
>
 "(T ? 4 : 5)"
          -
>
 "4"                                    -
>
 "4"

Example 3:

Input:
 "T?T?F:5:3"


Output:
 "F"


Explanation:
 The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(T ? (T ? F : 5) : 3)"                   "(T ? (T ? F : 5) : 3)"
          -
>
 "(T ? F : 3)"                 or       -
>
 "(T ? F : 5)"
          -
>
 "F"                                    -
>
 "F"

Hint

DFS + stack

Method

Firstly, expression problem seems like the calculator, maybe we can use stack

Secondly, each ternary expression including a mini ternary expression just like a recursion

so we can use dfs by stack

since it group from right to left

we can start from right

each time we meet problem sign we do a mini calculate

T?F?4:5:6

  do first cal -> F ? 4 : 5 -> 5

T?5:6

do second cal-> 5

Time & Space

O(n)

Code

public String parseTernary(String expression) {

if (expression == null || expression.length() == 0){

return "";

}

Deque<Character> stack = new LinkedList<Character>();

for (int i = expression.length() - 1; i >= 0; i--){

char c = expression.charAt(i);

if (!stack.isEmpty() && stack.peek() == '?'){

stack.pop();

char first = stack.pop();

stack.pop();

char second = stack.pop();

if (c == 'T'){

stack.push(first);

} else {

stack.push(second);

}

} else {

stack.push(c);

}

}

return stack.peek() + "";

}