題目出處
weekly-contest-282
2185. Counting Words With a Given Prefix
難度
Easy
個人範例程式碼
class Solution:
def prefixCount(self, words: List[str], pref: str) -> int:
pref_length = len(pref)
ans_cnt = 0
for ele in words:
if len(ele) < pref_length:
continue
else:
if self.compare_str(ele[0: pref_length], pref):
ans_cnt += 1
else:
pass
return ans_cnt
def compare_str(self, word_pref: str, pref: str) -> bool:
return (word_pref == pref)
Time Complexity
O(n)
算法說明
沒什麼,就取子字串直接比較
corner case 特殊情況處理
x
Boundary conditions/ Edge conditions 邊際情況處理
- python 子字串長度
str[0:n] 取得範圍為 str 的 index 0 到 index n-1(包含)