791. Custom Sort String

Use  char[] array 

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.




=======================================

class Solution {
    public String customSortString(String S, String T) {
        // store the char and freq from T that appears in S
        HashMap<Character, Integer> map1 = new HashMap<>();
        // store the char and freq from T that  didnt appear in S
        HashMap<Character, Integer> map2 = new HashMap<>(); ////// not char 
        // store chars of S 
        HashSet<Character> set = new HashSet<>();
            
        
        for(char c : S.toCharArray()){
            set.add(c);
        }
        
        for(char c : T.toCharArray()){
            if(set.contains(c)){
                Integer freq = map1.get(c); //// not int 
                if(freq == null){
                    map1.put(c, 1);
                }else{
                    map1.put(c, freq + 1);
                }
            }else{
                Integer freq = map2.get(c);
                if(freq == null){
                    map2.put(c, 1);
                }else{
                    map2.put(c, freq + 1);
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for(char c : S.toCharArray()){
            Integer freq = map1.get(c);
            if(freq != null){ ///// c might not be in map1 , there might be elements in S that is not in T
                while(freq > 0){
                sb.append(c);
                freq--;
                }
            }
        }
          
        for(char c : map2.keySet()){
            int freq = map2.get(c);
            while(freq > 0){
                sb.append(c);
                freq--;
            }
        }
        return sb.toString();
        
    }
}






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class Solution {
    public String customSortString(String S, String T) {
        StringBuilder sb = new StringBuilder();
        HashMap<Character, Integer> map = new HashMap<>();
        for(char c : T.toCharArray()){ // T.toCharArray()
            Integer freq = map.get(c);
            if(freq == null){
                map.put(c, 1);
            }else{
                map.put(c, freq + 1);
            }
        }
        
        for(char c : S.toCharArray()){
            Integer freq = map.get(c); // Integer, because might return null
            if(freq != null){
                while(freq > 0){
                    sb.append(c);
                    freq--;
                }
                map.remove(c);
            }
        }
        
        for(char c : map.keySet()){
            int freq = map.get(c);
            while(freq > 0){
                sb.append(c);
                freq--;
            }
        }
        return sb.toString();
        
    }
}

======================


class Solution {
    public String customSortString(String S, String T) {
        StringBuilder sb = new StringBuilder();
        int[] count = new int[26];
        for(char c : T.toCharArray()){
            count[c - 'a']++;
        }
        
        for(char c : S.toCharArray()){
            int freq = count[c - 'a'];
            if(freq > 0){
                while(freq > 0){
                    sb.append(c);
                    freq--;
                }
                count[c - 'a'] = 0;
            }
            
        }
        
        for(char c = 'a'; c <= 'z'; c++){
            int freq = count[c - 'a'];
            if(freq != 0){
                while (freq > 0){
                    sb.append(c);
                    freq--;
                }
            }
        }
        return sb.toString(); 
    }
}


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https://leetcode.com/problems/custom-sort-string/solution/

一会再看下这个答案