Question 139 Word Break

Link:

https://leetcode.com/problems/word-break/

My code:

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
            if len(s) ==0:
                return False
            elif len(s) ==1:
                if s[0] in wordDict:
                    return True
                else:
                    return False
            else:
                TFList  = [True]
                InitialString = s[0]
                if InitialString in wordDict:
                    TFList.append(True)
                else:
                    TFList.append(False)
                for i in range(1,len(s)):
                    ExitFlag = False
                    TFValue = TFList[i]
                    if TFValue == True:
                        String1 = s[i]
                        if String1 in wordDict:
                            TFList.append(True)
                            continue
                        else:
                            for j in range(1,(1+i)):
                                string2 = s[(i-j):i+1]
                                if TFList[i-j] ==True and string2 in wordDict:
                                    ExitFlag = True
                                    TFList.append(True)
                                    break
                        if ExitFlag == True:
                            continue
                        else:
                            TFList.append(False)
                    else:
                        for j in range(1,(i+1)):
                                string2 = s[(i-j):(i+1)]
                                if TFList[i-j] ==True and string2 in wordDict:
                                    ExitFlag = True
                                    TFList.append(True)
                                    break
                        if ExitFlag == True:
                            continue
                        else:
                            TFList.append(False)
                return TFList[len(s)]

Explanation: The idea of my code is to recurse from the start of the string. For each extra string we attach to original string, there are two situations: 1. Old string can be broken by the given dictionary. In this case, we first test if the extra one string is in the dictionary. If true, then we return True for this substring. If not, we need to go back to test every case with strings divided by i-j the element. If one of them gives true, then we may return true, otherwise give false for that case.