标签:not count pytho bre [] value 输入 and counter
示例:
输入:s = "abcmmedfrgaqwedfrmme" words=["mme","dfr"]
输出:[3,14]
Python解决方案:
class Solution(object): def findSubstring(self, s, words): """ :type s: str :type words: List[str] :rtype: List[int] """ if not words: return [] w_len = len(words[0]) all_len = w_len*len(words) out = [] word_count = collections.Counter(words) for i in range(len(s)-all_len+1): sub_i = s[i:i+w_len] if sub_i in word_count: j = i + w_len wc = word_count.copy() wc[sub_i] -= 1 while j <= all_len + i-w_len: sub = s[j:j+w_len] if sub in wc and wc[sub]: wc[sub] -= 1 j += w_len pre = sub else: break if not sum(wc.values()): out.append(i) return out
给定一个字符串s和一个由多个等长字符串组成的列表words,输出列表中的字符串组成的整体在s中的所有可能的位置
标签:not count pytho bre [] value 输入 and counter
原文地址:https://www.cnblogs.com/wenqinchao/p/10688959.html