传送门：784. Letter Case Permutation
Problem:
Given a string S, we can transform every letter individually to be lowercase or uppercase to create another string. Return a list of all possible strings we could create.
Examples:
Input: S = “a1b2” Output: [“a1b2”, “a1B2”, “A1b2”, “A1B2”]
Input: S = “3z4” Output: [“3z4”, “3Z4”]
Input: S = “12345” Output: [“12345”]
Note:
S will be a string with length at most 12.S will consist only of letters or digits.思路： dfs即可，遇到数字跳过，遇到字母，分为两种情况传给子问题。
Java版本：
public List<String> letterCasePermutation(String S) { all = new ArrayList<>(); backtrack(S.toCharArray(), 0, ""); return all; } List<String> all; public void backtrack(char[] cs, int pos, String ans) { if (pos == cs.length) { all.add(ans); return; } else { if (Character.isAlphabetic(cs[pos])) { char l = cs[pos]; backtrack(cs, pos + 1, ans + l); if (l >= 'a' && l <= 'z') l = (char) (l - 'a' + 'A'); else if (l >= 'A' && l <= 'Z') l = (char) (l - 'A' + 'a'); backtrack(cs, pos + 1, ans + l); } else { backtrack(cs, pos + 1, ans + cs[pos]); } } }Python版本：
class Solution(object): def letterCasePermutation(self, S): """ :type S: str :rtype: List[str] """ ans = [] def dfs(S, pos, str): if pos == len(S): ans.append(str) return else: if S[pos].isalpha(): letter = S[pos] dfs(S, pos + 1, str + letter.upper()) dfs(S, pos + 1, str + letter.lower()) else: dfs(S, pos + 1, str + S[pos]) dfs(S, 0, '') return ans