leetcode-30-串联所有单词的子串

题目描述：

 

 方法一：暴力O(MN)

class Solution:
    def findSubstring(self, s: str, words: List[str]) -> List[int]:
        res = []
        if not s or not words:
            return res
        for j in range(len(s)-len(words[0])*len(words)+1):
            i = j
            tmp = words.copy()
            while i - j <= (len(words)-1)*len(words[0]):
                if s[i:i+len(words[0])] in tmp:
                    tmp.remove(s[i:i+len(words[0])])
                    if not tmp:
                        res.append(j)
                    i += len(words[0])
                else:
                    break
        return res

优化：

class Solution:
    def findSubstring(self, s: str, words: List[str]) -> List[int]:
        from collections import Counter
        if not s or not words:return []
        one_word = len(words[0])
        all_len = len(words) * one_word
        n = len(s)
        words = Counter(words)
        res = []
        for i in range(0, n - all_len + 1):
            tmp = s[i:i+all_len]
            c_tmp = []
            for j in range(0, all_len, one_word):
                c_tmp.append(tmp[j:j+one_word])
            if Counter(c_tmp) == words:
                res.append(i)
        return res

方法二：滑动窗口： O(N)

class Solution:
    def findSubstring(self, s: str, words: List[str]) -> List[int]:
        from collections import Counter
        if not s or not words:return []
        one_word = len(words[0])
        word_num = len(words)
        n = len(s)
        if n < one_word:return []
        words = Counter(words)
        res = []
        for i in range(0, one_word):
            cur_cnt = 0
            left = i
            right = i
            cur_Counter = Counter()
            while right + one_word <= n:
                w = s[right:right + one_word]
                right += one_word
                if w not in words:
                    left = right
                    cur_Counter.clear()
                    cur_cnt = 0
                else:
                    cur_Counter[w] += 1
                    cur_cnt += 1
                    while cur_Counter[w] > words[w]:
                        left_w = s[left:left+one_word]
                        left += one_word
                        cur_Counter[left_w] -= 1
                        cur_cnt -= 1
                    if cur_cnt == word_num :
                        res.append(left)
        return res

 

来源：https://www.cnblogs.com/oldby/p/11771448.html