LeetCode Biweekly Contest No. 81 (1/4)-优快云博客

该博客介绍了如何解决一个字符串处理问题，即在给定的字符串中，统计每一对垂直条' '| '之间的'*'数量，但不包括特定分组的'*'。解题思路涉及遍历字符串，通过跟踪' '| '的奇偶性来确定哪些'*'应该被计数。给出的Python代码实现了一个高效解决方案，时间复杂度为O(N)，空间复杂度为O(1)。

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Problem 1. 2315. Count Asterisks

题目描述：

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 '|'.

解题思路：

问题是统计两个"|"（称为"|"对儿）之间的"*"的个数，并且排除掉一部分"|"对儿。

排除掉的"|"对儿的特征是：

第1、2个，第3、4个，第5，6个……

基于这样的特征，我们用一个bool变量b存储当前"|"的个数的奇偶性，当为偶数个时，对"*"进行计数。每次遇到新的"|"时，变更b的状态，于是：

class Solution:
    def countAsterisks(self, s: str) -> int:
        ret = 0
        b = False
        for i, t in enumerate(s):
            if t == '|':
                b = not b
            if not b and t == '*':
                ret += 1
        return ret

时间复杂度：O(N)

额外空间复杂度：O(1)