LeetCode 791. Custom Sort String

原题链接在这里：https://leetcode.com/problems/custom-sort-string/

题目：

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.

题解：

Mark each char and frequency in T with map.

Then for each cahr sc in S, check if it is in T as well. If yes then append it after res with frequency times. And remove it from map.

Then the rest in map are char in T, but not in S. Append them after res.

Time Complexity: O(m+n). m = S.length(). n = T.length().

Space: O(n).

AC Java:  

 1 class Solution {
 2     public String customSortString(String S, String T) {
 3         if(T == null || T.length() == 0 || S == null || S.length() == 0){
 4             return T;
 5         }
 6         
 7         int [] map = new int [26];
 8         for(char c : T.toCharArray()){
 9            map[c-'a']++;
10         }
11         
12         StringBuilder sb = new StringBuilder();
13         for(char sc : S.toCharArray()){
14             while(map[sc-'a'] > 0){
15                 sb.append(sc);
16                 map[sc-'a']--;
17             }
18         }
19         
20         for(char c = 'a'; c<='z'; c++){
21             while(map[c-'a']>0){
22                 sb.append(c);
23                 map[c-'a']--;
24             }
25         }
26         
27         return sb.toString();
28     }
29 }