題目

基本思路

使用一個字典統計一下words中每個單詞的數量。由於每個單詞的長度一樣，以題中給的例子而言，可以3個字母3個字母的檢查，如果不在字典中，則break出迴圈。有一個技巧是建立一個臨時字典currDict，用來統計s中那些在words中的單詞的數量，必須和words中單詞的數量相等，否則同樣break。

實現程式碼

class Solution(object):
    def findSubstring(self, s, words):
        """
        :type s: str
        :type words: List[str]
        :rtype: List[int]
        """
 

        wordsNum = len(words)
        sLen = len(s)
        
        if wordsNum == 0 or sLen == 0:
            return []
        
        wordLen = len(words[0])
        wordsDict = {}
        for word in words:
            wordsDict[word] = wordsDict[word]+1 if word in wordsDict else 1
        
        res =
 
 []
        
        # i 代表起始位置
        for i in range(sLen+1-wordsNum*wordLen):
            curDict = {}
            j = 0
            while j < wordsNum:
                word = s[i+j*wordLen:i+j*wordLen+wordLen]
                
                if word not in wordsDict:
                    break
 

                
                curDict[word] = curDict[word]+1 if word in curDict else 1
                
                if curDict[word] > wordsDict[word]:
                    break
                j += 1
                
            if j == wordsNum:
                res.append(i)
        
        return res