题目

题目即给出一个字符串，然后对数字不做处理，对字母进行排序组合，然后输出排序组合得到的列表。

纯字母大小写排序组合思路参考：

def strSort(words):
tempword = words
for i in range(2**len(words)):
int_bin = str(bin(i)).replace(“0b”,"")
for i in range(len(words)-len(int_bin)):
int_bin = “”.join((“0”,int_bin))
for index, i in enumerate(int_bin):
if i==‘1’:
wordslst = list(words)
wordslst[index] = wordslst[index].upper()
words = “”.join(wordslst)
yield words
words = tempword

for word in strSort(“def”):
想了挺久感觉题目比较难，仅仅考虑到了大小写排列思路怎么写。写出来结果会过于复杂。
于是查询了讨论区：

优秀代码（极其巧妙）

从前往后，没有使用递归，使用了for，效率略低

import itertools
class Solution(object):
    def letterCasePermutation(self, S):
        """
        :type S: str
        :rtype: List[str]
        """
        res = ['']
        for ch in S:
            if ch.isalpha():
                res = [i+j for i in res for j in [ch.upper(), ch.lower()]]
            else:
                res = [i+ch for i in res]
        return res

从后往前，使用了递归效率更高

class Solution(object):
  def letterCasePermutation(self, S):
      if len(S) == 0:
          return [""]
      temp_result = self.letterCasePermutation(S[1:])
      c = S[0]
      if c.isalpha():
          result = [ u+r for u in [c.upper(), c.lower()] for r in temp_result]
      else:
          result = [ c+r for r in temp_result ]       
      return result

同样是从后往前，思路没有之前的精巧

class Solution(object):
    def letterCasePermutation(self, S):
        return self.permute(S)
    def permute(self, str):
        lenOfStr = len(str)
        if lenOfStr == 0:
            return [""]
        lastElem = str[lenOfStr-1]
        str = str[:lenOfStr-1]
        tempAns = self.permute(str)
        ans = []
        for permutation in tempAns:
            if lastElem.isalpha():
                ans.append(permutation + lastElem.lower())
                ans.append(permutation + lastElem.upper())
            else:
                ans.append(permutation + lastElem)
        return ans