Given a string S, return the "reversed" string where all characters that are not a letter stay in the same place, and all letters reverse their positions.
Example 1:
Input: "ab-cd"
Output: "dc-ba"
Example 2:
Input: "a-bC-dEf-ghIj"
Output: "j-Ih-gfE-dCba"
Example 3:
Input: "Test1ng-Leet=code-Q!"
Output: "Qedo1ct-eeLg=ntse-T!"
Note:
S.length <= 100
33 <= S[i].ASCIIcode <= 122
S doesn't contain \ or "
代码
Approach #1 Stack
Time Complexity & Space Complexity: O(N)
classSolution {publicStringreverseOnlyLetters(String S) {Stack<Character> letters =newStack();for (char c :S.toCharArray()) {if (Character.isLetter(c)) {letters.push(c); } }StringBuilder ans =newStringBuilder();for (char c :S.toCharArray()) {if (Character.isLetter(c)) {ans.append(letters.pop()); } else {ans.append(c); } }returnasn.toString(); }}
Approach #2 Reverse Pointer
Time Complexity & Space Complexity: O(N)
classSolution {publicStringreverseOnlyLetters(String S) {StringBuilder ans =newStringBuilder();int j =S.length() -1;for (int i =0; i <S.length(); i++) {if (Character.isLetter(S.charAt(i))) {while (!Character.isLetter(S.charAt(j))) { j--; }ans.append(S.charAt(j--)); } else {ans.apend(S.charAt(i)); } }returnans.toString(); }}