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English Version

题目描述

请你设计一个可以解释字符串 commandGoal 解析器command"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 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"G""()" 和/或 "(al)" 按某种顺序组成

解法

方法一:字符串替换

根据题意,只需要将字符串 command 中的 "()" 替换为 'o'"(al)" 替换为 "al" 即可。

方法二:字符串遍历

我们也可以遍历字符串 command,对于每个字符 $c$

  • 如果是 'G',直接将 $c$ 添加到结果串中;
  • 如果是 '(',判断下一个字符是否是 ')',若是,将 'o' 添加到结果串中,否则,将 "al" 添加到结果串中。

遍历结束,返回结果串即可。

时间复杂度 $O(n)$,空间复杂度 $O(1)$

Python3

class Solution:
    def interpret(self, command: str) -> str:
        return command.replace('()', 'o').replace('(al)', 'al')
class Solution:
    def interpret(self, command: str) -> str:
        ans = []
        for i, c in enumerate(command):
            if c == 'G':
                ans.append(c)
            elif c == '(':
                ans.append('o' if command[i + 1] == ')' else 'al')
        return ''.join(ans)

Java

class Solution {
    public String interpret(String command) {
        return command.replace("()", "o").replace("(al)", "al");
    }
}
class Solution {
    public String interpret(String command) {
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < command.length(); ++i) {
            char c = command.charAt(i);
            if (c == 'G') {
                ans.append(c);
            } else if (c == '(') {
                ans.append(command.charAt(i + 1) == ')' ? "o" : "al");
            }
        }
        return ans.toString();
    }
}

C++

class Solution {
public:
    string interpret(string command) {
        while (command.find("()") != -1) command.replace(command.find("()"), 2, "o");
        while (command.find("(al)") != -1) command.replace(command.find("(al)"), 4, "al");
        return command;
    }
};
class Solution {
public:
    string interpret(string command) {
        string ans;
        for (int i = 0; i < command.size(); ++i) {
            char c = command[i];
            if (c == 'G') ans += c;
            else if (c == '(') ans += command[i + 1] == ')' ? "o" : "al";
        }
        return ans;
    }
};

Go

func interpret(command string) string {
    command = strings.ReplaceAll(command, "()", "o")
    command = strings.ReplaceAll(command, "(al)", "al")
    return command
}
func interpret(command string) string {
	ans := &strings.Builder{}
	for i, c := range command {
		if c == 'G' {
			ans.WriteRune(c)
		} else if c == '(' {
			if command[i+1] == ')' {
				ans.WriteByte('o')
			} else {
				ans.WriteString("al")
			}
		}
	}
	return ans.String()
}

TypeScript

function interpret(command: string): string {
    return command.replace(/\(\)/g, 'o').replace(/\(al\)/g, 'al');
}
function interpret(command: string): string {
    const n = command.length;
    const ans: string[] = [];
    for (let i = 0; i < n; i++) {
        const c = command[i];
        if (c === 'G') {
            ans.push(c);
        } else if (c === '(') {
            ans.push(command[i + 1] === ')' ? 'o' : 'al');
        }
    }
    return ans.join('');
}

Rust

impl Solution {
    pub fn interpret(command: String) -> String {
        command.replace("()", "o").replace("(al)", "al")
    }
}
impl Solution {
    pub fn interpret(command: String) -> String {
        let mut ans = String::new();
        let bs = command.as_bytes();
        for i in 0..bs.len() {
            if bs[i] == b'G' {
                ans.push_str("G");
            }
            if bs[i] == b'(' {
                ans.push_str({
                    if bs[i + 1] == b')' {
                        "o"
                    } else {
                        "al"
                    }
                })
            }
        }
        ans
    }
}

C

char *interpret(char *command) {
    int n = strlen(command);
    char *ans = malloc(sizeof(char) * n + 1);
    int i = 0;
    for (int j = 0; j < n; j++) {
        char c = command[j];
        if (c == 'G') {
            ans[i++] = 'G';
        } else if (c == '(') {
            if (command[j + 1] == ')') {
                ans[i++] = 'o';
            } else {
                ans[i++] = 'a';
                ans[i++] = 'l';
            }
        }
    }
    ans[i] = '\0';
    return ans;
}

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