LeetCode（String)2315. Count Asterisks

冷不防 2023-09-27 20:53 128阅读 0赞

# 1.问题 #

You are given a string s, where every two consecutive vertical bars ‘|’ are grouped into a pair. In other words, the 1st and 2nd ‘|’ make a pair, the 3rd and 4th ‘|’ make a pair, and so forth.

Return the number of ‘*’ in s, excluding the '*’ between each pair of ‘|’.

Note that each ‘|’ will belong to exactly one pair.

Example 1:

> Input: s = “l|*e*et|c**o|\*de|"  
> Output: 2  
> Explanation: The considered characters are underlined: "l|*e*et|c**o|\*de|”.  
> The characters between the first and second ‘|’ are excluded from the answer.  
> Also, the characters between the third and fourth ‘|’ are excluded from the answer.  
> There are 2 asterisks considered. Therefore, we return 2.

Example 2:

> Input: s = “iamprogrammer”  
> Output: 0  
> Explanation: In this example, there are no asterisks in s. Therefore, we return 0.

Example 3:

> Input: s = “yo|uar|e\*\*|b|e***au|tifu|l"  
> Output: 5  
> Explanation: The considered characters are underlined: "yo|uar|e**|b|e*\*\*au|tifu|l”. There are 5 asterisks considered. Therefore, we return 5.

Constraints:

1 <= s.length <= 1000  
s consists of lowercase English letters, vertical bars ‘|’, and asterisks ‘\*’.  
s contains an even number of vertical bars ‘|’.

# 2. 解题思路 #

## 方法1： ##

> 1.定义"|" 和"*"的计数器  
> 2.将字符串转换为字符数组  
> 3.如果当前字符是 “|”，将 “|” 的计数器加1  
> 4.如果 “|” 的数量是偶数，统计 "*" 的数量并且当前字符是 “*"，将 "*” 的计数器加1  
> 5.返回 “\*” 的计数器

## 方法2： ##

> 1.标记当前字符是否在 “|” 内部  
> 2.统计在偶数个 “|” 之间的 “*" 的数量  
> 3.如果当前字符是 “|”，则切换 “insidePipe” 的状态  
> 4.如果当前字符不在 “|” 内部，并且是 "*”，则将 “*" 的计数器加1  
> 5.返回 "*” 的计数器

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public int countAsterisks(String s) {
        
            int bar_count = 0; // "|" 的计数器
            int asterisk_count = 0; // "*" 的计数器
            char[] char_arr = s.toCharArray(); // 将字符串转换为字符数组
            // 遍历字符串
            for (int i = 0; i < s.length(); i++) {
        
                if(char_arr[i] == '|'){
        
                    bar_count++; // 如果当前字符是 "|"，将 "|" 的计数器加1
                }
                // 如果 "|" 的数量是偶数，统计 "*" 的数量
                if(bar_count % 2 ==0&&char_arr[i]  == '*'){
        
    				 asterisk_count++; // 如果当前字符是 "*"，将 "*" 的计数器加1           
                }
            }
            return asterisk_count; // 返回 "*" 的计数器
        }
    }

## 代码2： ##

class Solution {
        
        public int countAsterisks(String s) {
        
            boolean insidePipe = false; // 1.标记当前字符是否在 "|" 内部
            int count = 0; // 2.统计在偶数个 "|" 之间的 "*" 的数量
            for(int i = 0; i < s.length(); i++){
        
                if(s.charAt(i) == '|'){
        
                    insidePipe = !insidePipe; // 3.如果当前字符是 "|"，则切换 "insidePipe" 的状态
                }
                if(!insidePipe && s.charAt(i) == '*'){
        
                    count++; // 4.如果当前字符不在 "|" 内部，并且是 "*"，则将 "*" 的计数器加1
                }
            }
            return count; // 5.返回 "*" 的计数器
        }
    }