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|eet|co|*de|" Output: 2 Explanation: The considered characters are underlined: "l|eet|co|*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|eau|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
    }
}