[LeetCode] 791. Custom Sort String

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.

题意：给字符串S和T，把T中的部分(含有S的部分)序列按照S的序列排序，剩余的接到尾部

S中没有重复的字符，S和T中也只有小写字母，总共26个字母，桶排序或者计数排序都可以；

class Solution {
    public String customSortString(String S, String T) {
        int[] str = new int [26];
        for (int i = 0; i < T.length(); i++)
            str[T.charAt(i) - 'a'] ++;
        String strs = "";
        for (int i= 0; i < S.length(); i++) {
            if (str[S.charAt(i) - 'a'] != 0) {
                for (int j = 0; j < str[S.charAt(i) - 'a']; j++) {
                    strs += S.charAt(i);
                }
                str[S.charAt(i) - 'a'] = 0;
            }
        }
        for (int i = 0; i < 26; i++) {
            if (str[i] != 0) {
                for (int j = 0; j < str[i]; j++) {
                    strs += (char)(i + 'a');
                }
            }
        }
        return strs;
    }
}