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);
     }
 }