439. 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 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"

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"

这道题是三元表达式解析程序，我的思路是从后面开始遍历，遇到？就求解答案，然后再把答案赋值到？之前，接着把这个式子删除，然后继续寻找？直达字符串只剩一个字符。

class Solution {public:    string parseTernary(string expression)     {        while(expression.length()>1)        {            int i=expression.length()-1;            while(expression[i]!='?')                i--;            char c;            if(expression[i-1]=='T')                c=expression[i+1];            else                c=expression[i+3];            expression[i-1]=c;            expression.erase(i,4);        }        return expression;    }};