Leetcode Biweekly Contest 81

發表於 更新於 分類於

Graph 題我直接當場去世。

2315. Count Asterisks

Difficulty: Easy, Open in Leetcode

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.

題目解釋

以 | 為分隔下去 split，算出奇數 index 的 * 有幾個

Solution

不知道算不算暴力。先用 split() 依照 | 分開字串，接著用 count() 計算奇數 index 中的 *，
加總後就是答案了。

1
2
3
4
5
6
7
8
9
class Solution:
    def countAsterisks(self, s: str) -> int:
        s_arr = s.split("|")
        n = len(s_arr)
        result = 0
        for i in range(0, n, 2):
            result = result + s_arr[i].count("*")
            
        return result

來看另一種解法，跑過一次的同時紀錄 | 狀況，最後也是加總就答案。

1
2
3
4
5
6
7
8
9
10
11
def countAsterisks(self, s: str) -> int:
    bar = False
    count = 0
    for x in s:
        if x=='|':
            bar = not bar
                
        if x=='*' and bar==False:
            count+=1
            
    return count