博客
关于我
Leetcode: Ternary Expression Parser
阅读量:806 次
发布时间:2023-01-31

本文共 3374 字,大约阅读时间需要 11 分钟。

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
st = new Stack
(); 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
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 }

 

转载地址:http://gmgyk.baihongyu.com/

你可能感兴趣的文章
MySQL-redo日志
查看>>
MySQL-【1】配置
查看>>
MySQL-【4】基本操作
查看>>
Mysql-丢失更新
查看>>
Mysql-事务阻塞
查看>>
Mysql-存储引擎
查看>>
mysql-开启慢查询&所有操作记录日志
查看>>
MySQL-数据目录
查看>>
MySQL-数据页的结构
查看>>
MySQL-架构篇
查看>>
MySQL-索引的分类(聚簇索引、二级索引、联合索引)
查看>>
Mysql-触发器及创建触发器失败原因
查看>>
MySQL-连接
查看>>
mysql-递归查询(二)
查看>>
MySQL5.1安装
查看>>
mysql5.5和5.6版本间的坑
查看>>
mysql5.5最简安装教程
查看>>
mysql5.6 TIME,DATETIME,TIMESTAMP
查看>>
mysql5.6.21重置数据库的root密码
查看>>
Mysql5.6主从复制-基于binlog
查看>>