leetcode: 30. Substring with Concatenation of All Words [✗]

最新推荐文章于 2025-08-04 20:35:07 发布

JNingWei

最新推荐文章于 2025-08-04 20:35:07 发布

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分类专栏： LeetCode-Python 文章标签： leetcode

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本文介绍了一种高效的字符串匹配算法，该算法能在给定的字符串中找到由多个指定单词组成的子串的所有起始位置。通过使用Python实现的具体案例，展示了如何利用哈希表来优化查找过程。

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Problem

# You are given a string, s, and a list of words, words, that are all of the same length.
# Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
#
# For example, given:
# s: "barfoothefoobarman"
# words: ["foo", "bar"]
#
# You should return the indices: [0,9].
# (order does not matter).

AC

class Solution():
    def findSubstring(self, s, words):
        import collections
        dict  = collections.defaultdict(int)
        for w in words:
            dict[w] += 1
        res, l = [], len(words[0])
        for i in range(l):
            dict1, j = dict.copy(), i
            while j < len(s) - l + 1:
                dict1[s[j:j+l]] -= 1
                while dict1[s[j:j+l]] < 0:
                    dict1[s[i:i+l]] += 1
                    i += l
                j += l
                if (j-i) / l == len(words) :
                    res.append(i)
        return res


if __name__ == "__main__":
    assert Solution().findSubstring("barfoothefoobarman", ["foo", "bar"]) == [0, 9]