[Leetcode] Ternary Expression Parser, Solution

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:

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"

[Thoughts]

从右往左扫，发现？号就处理掉，最后的结果就是。


[Code]

1:    string parseTernary(string expression) {  
2:      int len = expression.size();  
3:        
4:      if(len == 0) return expression;  
5:        
6:      for(int i = len-1; i>=0; i--) {  
7:        if(expression[i] != '?') continue;  
8:          
9:        int index = i;  
10:        char result;  
11:        if(expression[index-1] == 'T'){  
12:          result = expression[index+1];  
13:        }else {  
14:          result = expression[index+3];  
15:        }  
16:          
17:        expression.erase(index -1, 5);  
18:        expression.insert(index-1, 1, result);  
19:      }  
20:      return expression;  
21:    }