(LeetCode) Custom Sort String - LeetCode

Description:

S and T are strings composed of lowercase letters. In S, no letter occurs more than once.

S was sorted in some custom order previously. We want to permute the characters of T so that they match the order that S was sorted. More specifically, if x occurs before y in S, then x should occur before y in the returned string.

Return any permutation of T (as a string) that satisfies this property.

Example :
Input: 
S = "cba"
T = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.

Note:

S has length at most 26, and no character is repeated in S.
T has length at most 200.
S and T consist of lowercase letters only.

Accepted

40K

Submissions

63.5K

Solution:

There are two helper functions I used.

1. String charRemoveAt(String str, int p) ;

with the help of this function, I can delete character at position p in the given String str.

2. Sort a String

Method 1(natural sorting) :

Apply toCharArray() method on input string to create a char array for input string.
Use Arrays.sort(char c[]) method to sort char array.
Use String class constructor to create a sorted string from char array.

class Solution {
    public String customSortString(String S, String T) {
        
        if(S==null||S.length()==0||T==null||T.length()==0){
            return null;
        }
        
        
        String res = "";
        
        for(int i = 0; i<S.length(); i++){
            
           // System.out.println(S.charAt(i));
           while (T.indexOf(S.charAt(i))>=0){
                
                res = res + Character.toString((S.charAt(i)));
                                               
                T = charRemoveAt(T,T.indexOf(S.charAt(i)));
            }
        }
        
         char tempArray[] = T.toCharArray();
         Arrays.sort(tempArray); 
         T = new String(tempArray);
        
         //System.out.println(charRemoveAt(T,T.indexOf('e')));
        return res + T;
        
        
    }
    
      public static String charRemoveAt(String str, int p) {  
              return str.substring(0, p) + str.substring(p + 1);  
           }  
}

原文地址：https://www.cnblogs.com/codingyangmao/p/11451219.html