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 digits 0-9, ?, :, T and F (T and F represent True and False respectively).
Note:
- The length of the given string is ≤ 10000.
- Each number will contain only one digit.
- The conditional expressions group right-to-left (as usual in most languages).
- The condition will always be either T or F. That is, the condition will never be a digit.
- 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"
Basic Idea:
Use Stack to process the whole String in a backward way. We process from back, to check the mark”?”, if we meet “?”, we take a look at the next condition, if true, we choose the result closed to the “?”, if not, we choose the another one.
class Solution{
public String parseTernary(String expression) {
if (expression == null || expression.length() == 0) return "";//check corner cases.
Deque<Character> stack = new LinkedList<>();
for (int i = expression.length() - 1; i >= 0; i--) {
char c = expression.charAt(i);
if (!stack.isEmpty() && stack.peek() == '?') {
stack.pop(); //pop '?'
char first = stack.pop();//the right answer
stack.pop(); //pop ':'
char second = stack.pop();//the false answer
if (c == 'T') stack.push(first);
else stack.push(second);
} else {
stack.push(c);
}
}
return String.valueOf(stack.peek());
}
}