LeetCode 439. Ternary Expression Parser

原题链接在这里：https://leetcode.com/problems/ternary-expression-parser/description/

题目：

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"

题解：

从后往前扫输入的string. 把char 放入stack中. 当扫过了'?', 就知道需要开始判断了.

从stack中弹出两个数，判断后的数放回stack中.

Time Complexity: O(n). n = expression.length();

Space: O(n).

AC Java:

class Solution {
    public String parseTernary(String expression) {
        if(expression == null || expression.length() == 0){
            return expression;
        }
        
        Stack<Character> stk = new Stack<Character>();
        for(int i = expression.length()-1; i>=0; i--){
            char c = expression.charAt(i);
            if(!stk.isEmpty() && stk.peek()=='?'){
                stk.pop(); // '?'
                char first = stk.pop();
                stk.pop(); // ':'
                char second = stk.pop();
                
                if(c == 'T'){
                    stk.push(first);
                }else{
                    stk.push(second);
                }
            }else{
                stk.push(c);
            }
        }
        
        return String.valueOf(stk.peek());
    }
}

Track the first string and second string separated by the corresponding " : ".

Based on the true or false, continue DFS on the corresponding string.

Time Complexity: O(n^2). n = expression.length().

Space: O(n).

AC Java: 

class Solution {
    public String parseTernary(String expression) {
        if(expression.length() == 1){
            return expression;
        }
        
        int indexQuestion = expression.indexOf("?");
        boolean flag = expression.charAt(0) == 'T' ? true : false;
        
        int count = 0;
        int start = indexQuestion+1;
        while(expression.charAt(start) != ':' || count != 0){
            if(expression.charAt(start) == '?'){
                count++;
            }else if(expression.charAt(start) == ':'){
                count--;
            }
            
            start++;
        }
        
        String first = expression.substring(indexQuestion+1, start);
        String second = expression.substring(start+1);
        return parseTernary(flag ? first : second);
    }
}