784. Letter Case Permutation---back tracking

题意： 字母变成大小写，数字不变Examples:
Input: S = "a1b2"
Output: ["a1b2", "a1B2", "A1b2", "A1B2"]

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

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

S = a1bc, 画出递归数如下：如果是字母 就是一个二叉树， 如果是数字就一个节点。所以本质上是一个二叉树的back tracking.

因为不熟悉 Java 字符串处理的一些function, 一开始写了一个特别丑陋的code:

class Solution {
    public List<String> letterCasePermutation(String S) {

        List<String> result = new ArrayList<>();
        StringBuilder sb = new StringBuilder(S);
        dfs(new StringBuilder(), sb, result,0);
        return result;

    }

    private void dfs(StringBuilder curResult, StringBuilder S, List<String> result, int index){

          if(curResult.length() == S.length()){
              result.add(curResult.toString());
              return;
          }

          char c = S.charAt(index);
          curResult.append(c);
          dfs(curResult,S,result,index+1);
          curResult.setLength(curResult.length()-1);   

          if(c>=‘a‘ && c<=‘z‘){

             char ch = change_char(c, true);
             S.setCharAt(index,ch);
             curResult.append(ch);
             dfs(curResult,S,result,index+1);
             curResult.setLength(curResult.length()-1);
             S.setCharAt(index,c);
           }  

           else if(c>=‘A‘ && c<=‘Z‘){
              char ch = change_char(c,false);
              S.setCharAt(index,ch);
              curResult.append(ch);
              dfs(curResult,S,result,index+1);
              curResult.setLength(curResult.length()-1);
              S.setCharAt(index,c);
          }
    }

    private char change_char(char c, boolean small){
        if(small){
            return (char)(c + ‘A‘-‘a‘);
        }
        else return  (char)(c + ‘a‘-‘A‘);
    }
}

用了Character 类里的function 后的code:

优化到了95%

class Solution {
    public List<String> letterCasePermutation(String S) {

        List<String> result = new ArrayList<>();
        dfs(new StringBuilder(), S, result,0);
        return result;

    }

    private void dfs(StringBuilder curResult, String S, List<String> result, int index){

          if(curResult.length() == S.length()){
              result.add(curResult.toString());
              return;
          }

          Character c = S.charAt(index); 

          if(Character.isLetter(c)){
              // left sub tree
              curResult.append(Character.toLowerCase(c));
              dfs(curResult,S,result,index+1);
              curResult.deleteCharAt(curResult.length()-1);

              //right sub tree
              curResult.append(Character.toUpperCase(c));
              dfs(curResult,S,result,index+1);
              curResult.deleteCharAt(curResult.length()-1);
          }

          else{ // is number
              curResult.append(c);
              dfs(curResult,S,result,index+1);
              curResult.deleteCharAt(curResult.length()-1);
          }
    }
}

原文地址：https://www.cnblogs.com/keepAC/p/9938463.html