#leetCode brush title documentary Day29

https://leetcode-cn.com/problems/letter-case-permutation/

Given a string S, S by the string shift each letter case, we can get a new string. Returns the set of all possible strings obtained.

Example:
Input: S = "a1b2"
Output: [ "a1b2", "a1B2 ", "A1b2", "A1B2"]

Input: S = "3z4"
Output: [ "3z4", "3Z4 "]

Input: S = "12345"
Output: [ "12345"]
Note:

Length S no more than 12.
S only by numbers and letters.

Chicken dishes to try:

The first feeling is back. I started out to write, but can not think of a recursive end condition of emmm

Worship Gangster Code:

[Recursive backtracking]

With vector <string> res store the results, s string to be processed

Recursion End Condition:
when S is an empty string, the end of the recursion, write result
recursive equatorial three cases:

Encountered a number, add this number on the letter continues recursively
encountered lowercase letters, lowercase letters on the letter added this continues recursively + Add this capital letters on the letter continues recursively
encountered in uppercase letters, add the capital letters on the letter recursively + Add to this the letter lowercase letters recursively

 1 class Solution {
 2 public:
 3     vector<string> res;
 4     vector<string> letterCasePermutation(string S) {
 5         letterCase("", S);
 6         return res;
 7     }
 8     void letterCase(string letter, string S) {
 9         if (S.size()==0) {
10             res.push_back(letter);
11             return;
12         }
13         char s = S[0];
14         letterCase(letter+s, S.substr(1));
15         if (s >= 'a' && s <='z') {
16             letterCase(letter + char(toupper(s)), S.substr(1));
17         } else if (s >= 'A' && s <='Z') {
18             letterCase(letter + char(tolower(s)), S.substr(1));
19         }
20     }
21 };

Taurus [O (n)]

. 1  class Solution {
 2  public :
 . 3      Vector < String > letterCasePermutation ( String S) {
 . 4          int size = s.size ();
 . 5          Vector < String > V;
 . 6          // initialization 
. 7          int NUM = . 1 ;
 . 8          v.push_back ( "" );
 . 9          for ( int I = 0 ; I <size; I ++) { // individually determined for each character string 
10              IF (the isalpha (S [I])) {// character 
. 11                  for ( int J = 0 ; J <NUM; J ++ ) {
 12 is                      String temp1 = V [J];
 13 is                      String temp2 of = temp1;
 14                      temp1 + = S [I];
 15                      V [J] = temp1 ;
 16                      S [I] ^ = 32 ; // clever use of the XOR operation, the conversion to complete the case 
. 17                      temp2 of + = S [I];
 18 is                      v.push_back (temp2 of);
 . 19                  }
 20 is                  NUM = NUM *2;
21             } else { // 是数字 
22                 for (int j = 0; j < num; j ++) {
23                     string temp = v[j];
24                     temp += S[i];
25                     v[j] = temp;
26                 }
27             }
28         }
29         return v;
30     }
31 };

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/letter-case-permutation
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

#leetCode brush title documentary Day29

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