Leetcode: Ternary Expression Parser-优快云博客

本文介绍了一种高效解析任意嵌套三元表达式的算法，使用栈结构避免了字符串拼接和子串操作，显著提高了运行效率。通过两个具体实现方案，详细展示了如何逐字符逆向遍历输入字符串，并正确处理条件表达式的右结合特性。

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"

My First Solution:

Use Stack and String operation, from the back of the string, find the first '?', push the right to stack. Depends on whether the char before '?' is 'T' or 'F', keep the corresponding string in the stack

 1 public class Solution {
 2     public String parseTernary(String expression) {
 3         Stack<String> st = new Stack<String>();
 4         int pos = expression.lastIndexOf("?");
 5         while (pos > 0) {
 6             if (pos < expression.length()-1) {
 7                 String str = expression.substring(pos+1);
 8                 String[] strs = str.split(":");
 9                 for (int i=strs.length-1; i>=0; i--) {
10                     if (strs[i].length() > 0)
11                         st.push(strs[i]);
12                 }
13             }
14             String pop1 = st.pop();
15             String pop2 = st.pop();
16             if (expression.charAt(pos-1) == 'T') st.push(pop1);
17             else st.push(pop2);
18             expression = expression.substring(0, pos-1);
19             pos = expression.lastIndexOf("?");
20         }
21         return st.pop();
22     }
23 }

Better solution, refer to https://discuss.leetcode.com/topic/64409/very-easy-1-pass-stack-solution-in-java-no-string-concat/2

No string contat/substring operation

 1 public String parseTernary(String expression) {
 2     if (expression == null || expression.length() == 0) return "";
 3     Deque<Character> stack = new LinkedList<>();
 4 
 5     for (int i = expression.length() - 1; i >= 0; i--) {
 6         char c = expression.charAt(i);
 7         if (!stack.isEmpty() && stack.peek() == '?') {
 8 
 9             stack.pop(); //pop '?'
10             char first = stack.pop();
11             stack.pop(); //pop ':'
12             char second = stack.pop();
13 
14             if (c == 'T') stack.push(first);
15             else stack.push(second);
16         } else {
17             stack.push(c);
18         }
19     }
20 
21     return String.valueOf(stack.peek());
22 }