统计星号【LC2315】

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.

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• 思路：记录每个'|'的编号，统计第偶数个'|'之后、第奇数个'|'之前的星号，以及末尾剩余字符串的星号个数

• 实现

class Solution {
    public int countAsterisks(String s) {
        int ans = 0;
        int count = 0;
        int idx = 0;
        for (char c : s.toCharArray()){
            if (c == '*'){
                count++;
            }else if(c == '|'){
                idx++;
                if (idx % 2 == 1){
                    ans += count;                    
                }
                count = 0;
            }
        }
        return ans + count;
    }
}

。复杂度

• 时间复杂度：O ( n ) ，n 为字符串的长度
• 空间复杂度：O ( 1 )