1678. 设计 Goal 解析器

淡淡的烟草味﹌ 2022-09-06 00:24 100阅读 0赞

[https://leetcode-cn.com/problems/goal-parser-interpretation/][https_leetcode-cn.com_problems_goal-parser-interpretation]

难度简单18

请你设计一个可以解释字符串 `command` 的 **Goal 解析器** 。`command` 由 `"G"`、`"()"` 和/或 `"(al)"` 按某种顺序组成。Goal 解析器会将 `"G"` 解释为字符串 `"G"`、`"()"` 解释为字符串 `"o"` ，`"(al)"` 解释为字符串 `"al"` 。然后，按原顺序将经解释得到的字符串连接成一个字符串。

给你字符串 `command` ，返回* ***Goal* *解析器 **对* *`command` 的解释结果。

**示例 1：**

输入：command = "G()(al)"
    输出："Goal"
    解释：Goal 解析器解释命令的步骤如下所示：
    G -> G
    () -> o
    (al) -> al
    最后连接得到的结果是 "Goal"

**示例 2：**

输入：command = "G()()()()(al)"
    输出："Gooooal"

**示例 3：**

输入：command = "(al)G(al)()()G"
    输出："alGalooG"

**提示：**

*  `1 <= command.length <= 100`
 *  `command` 由 `"G"`、`"()"` 和/或 `"(al)"` 按某种顺序组成

class Solution {
        public String interpret(String command) {
            String s = "";
            int n = 0;
            for(int i=0;i<command.length();i+=n)
            {
                if(command.charAt(i)=='G')
                {
                    n=1;
                    s+=""+command.charAt(i);
                }
                else if(command.charAt(i)=='(' && command.charAt(i+1)==')')
                {
                    n=2;
                    s+="o";
                }
                else if(command.charAt(i)=='(' && command.charAt(i+1)=='a')
                {
                    n=4;
                    s+="al";
                }
            }
            return s;
        }
    }

class Solution {
        public String interpret(String command) {
            command = command.replace("()","o");
           return  command.replace("(al)","al");
    
        }
    }

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[https_leetcode-cn.com_problems_goal-parser-interpretation]: https://leetcode-cn.com/problems/goal-parser-interpretation/
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