假设我们有一个由不同字符组成的字符串 s ,并且还有一个名为 words 的字符串数组。当字符串中的所有字符都出现在字符串 s 中时,该字符串是一致的。我们必须找到数组单词中存在的一致字符串的数量。
因此,如果输入像 s= "px", words = ["ad","xp","pppx","xpp","apxpa"],那么输出将是 3,因为只有三个字符串'p' 和 'x', ["xp","pppx","xpp"]。
让我们看看以下实现以获得更好的理解 -
def solve(s, words): count = 0 for i in range(len(words)): for j in range(len(words[i])): if words[i][j] not in s: break else: count += 1 return count s= "px" words = ["ad","xp","pppx","xpp","apxpa"] print(solve(s, words))
"px", ["ad","xp","pppx","xpp","apxpa"]输出结果
3