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Ternary Expression Parser
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
orF
. That is, the condition will never be a digit. - The result of the expression will always evaluate to either a digit
0-9
,T
orF
.
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"
1 public class Solution { 2 public String parseTernary(String expression) { 3 if (expression == null || expression.length() == 0) 4 return ""; 5 Stack<Character> stack = new Stack<>(); 6 7 for (int i = expression.length() - 1; i >= 0; i--) { 8 char c = expression.charAt(i); 9 if (!stack.isEmpty() && stack.peek() == ‘?‘) { 10 11 stack.pop(); // pop ‘?‘ 12 char first = stack.pop(); 13 stack.pop(); // pop ‘:‘ 14 char second = stack.pop(); 15 16 if (c == ‘T‘) 17 stack.push(first); 18 else 19 stack.push(second); 20 } else { 21 stack.push(c); 22 } 23 } 24 return String.valueOf(stack.peek()); 25 } 26 }
Ternary Expression Parser
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