Leetcode 30. Substring with Concatenation of All Words_1654. 出现次数最多的字母cat-only-iconcat 专属题目中文english给定一个字

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.

Example 1:

Input:
s = “barfoothefoobarman”,
words = [“foo”,“bar”]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are “barfoor” and “foobar” respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:

Input:
s = “wordgoodgoodgoodbestword”,
words = [“word”,“good”,“best”,“word”]
Output: []

思路

从字符串中检测出连续出现的单词表中的任意组合，返回起始位置的索引
可以用到滑动窗口的思想，这个思想解释起来有点抽象，
就相当于一个小的邻域范围内来进行搜索，匹配则记录。
窗口的大小是单词表的长度乘以每个相同长度的单词的长度，
需要移动窗口的步数是字符串的长度减去窗宽加上1 ，
每个窗口下都检测当前的字符是否在单词表
因为需要匹配单词表中的所有的单词，
所以只要不匹配就跳出当前循环。

code

class Solution:
    def findSubstring(self, s, words):
        if not words:
            return []
        n, wlen, slen, res = len(words), len(words[0]), len(s), []
        # 需要滑动窗口的次数, 总共需要检测连续的字母是 wlen * n
        slide = slen - wlen * n + 1
        for i in range(slide):
            temp = words.copy()
            for j in range(n):
                # 每次检测的字符存储在cur_word中
                cur_word = s[i + j * wlen:i + j * wlen + wlen]
                if cur_word in temp:
                    temp.remove(cur_word)
                else:
                    break
            if not temp:
                res.append(i)
        return res
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