leetcode 784. 字母大小写全排列(python)

￡神魔★判官ぃ 2022-04-22 11:14 200阅读 0赞

给定一个字符串`S`，通过将字符串`S`中的每个字母转变大小写，我们可以获得一个新的字符串。返回所有可能得到的字符串集合。

示例:
    输入: S = "a1b2"
    输出: ["a1b2", "a1B2", "A1b2", "A1B2"]
    
    输入: S = "3z4"
    输出: ["3z4", "3Z4"]
    
    输入: S = "12345"
    输出: ["12345"]

**注意：**

*  `S` 的长度不超过`12`。
 *  `S` 仅由数字和字母组成。

class Solution(object):
        def spliteList(self, alist):
            flag = 0    # 判断是否还含有两个字母的元素
            for i in range(len(alist)):
                k = alist[i]
                if len(k) == 2:
                    flag = 1
                    list1 = alist[:i] + [k[0].lower()] + alist[i+1:]
                    list2 = alist[:i] + [k[0].upper()] + alist[i+1:]
                    break
    
            if flag == 0:
                return alist
            else:
                return list1, list2
    
    
        def letterCasePermutation(self, S):
            """
            :type S: str
            :rtype: List[str]
            """
            sList = list( S )
            letterNum = 0
            for i in range(len(S)):
                k = sList[i]
                if 'A'<= k <= 'z':
                    letterNum += 1  # 统计一下字母的个数
                    sList[i] = k.lower() + k.upper()
    
            allList = []
            allList.append( sList )
    
            while len( allList ) != 2**letterNum:
                l = len( allList )
                for i in range(l):
                    list1 = allList[i]
                    if len( list1 ) != len( ''.join(list1)):
                        del allList[i]
                        l1, l2 = self.spliteList( list1 )
                        allList.append( l1 )
                        allList.append( l2 )
    
            l = len( allList )
            for i in range(l):
                allList[i] = ''.join( allList[i] )
    
            return allList