439 Ternary Expression Parser

439. Ternary Expression Parser

1. Question

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 eitherTorF. That is, the condition will never be a digit.

5. The result of the expression will always evaluate to either a digit0-9,TorF.

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"

2. Implementation

(1) Stack

思路：这是当年面Pocket Gem的电面题，当时的我完全懵逼无从入手...这道题既可以用递归，也可以用stack解。如果用stack的话，需要从后往前扫，这样的好处是当我们遇到'?'前面的character(T 或者 F)，我们可以知道该保留‘?’后面的哪个expression

class Solution {
    public String parseTernary(String expression) {
        if (expression == null || expression.length() == 0) {
            return "";
        }

        Stack<Character> stack = new Stack<>();

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

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

                char firstVal = stack.pop();
                stack.pop();
                char secondVal = stack.pop();

                if (c == 'T') {
                    stack.push(firstVal);
                }
                else {
                    stack.push(secondVal);
                }
            }
            else {
                stack.push(c);
            }
        }
        return String.valueOf(stack.pop());
    }
}

(2) DFS

class Solution {
    public String parseTernary(String expression) {
        int len = expression.length();

        if (len == 1) {
            return expression;
        }

        int count = 0;
        for (int i = 0; i < len; i++) {
            if (expression.charAt(i) == '?') {
                ++count;
            }
            else if (expression.charAt(i) == ':') {
                --count;

                // When the number of '?' matches with ':', we get a valid expression
                if (count == 0) {
                    if (expression.charAt(0) == 'T') {
                        // Get the expression before ':', skip 'T' and '?' at the beginning
                        return parseTernary(expression.substring(2, i));
                    } 
                    else {
                        // Get the expression after ':'
                        return parseTernary(expression.substring(i + 1));
                    }
                }
            }
        }
        return "";
    }
}

3. Time & Space Complexity

Stack: 时间复杂度O(n), 空间复杂度O(n)

DFS: 时间复杂度O(n), 空间复杂度O(n)