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:
- 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
TorF. That is, the condition will never be a digit. - The result of the expression will always evaluate to either a digit
0-9,TorF.
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);
}
}