791. 自定义字符串排序

解法一

根据S中各个字母出现的次数，定义字母优先级，然后进行自定义排序。

import java.util.Arrays;
import java.util.Comparator;
class Solution {
    // 字母优先级，越小优先级越高
    int[] priority = new int[26];
    public String customSortString(String S, String T) {
        int i;
        for (i = 0; i < S.length(); ++i) {
            priority[S.charAt(i) - 97] = i;
        }
        Comparator<Character> letterComparator = new Comparator<Character>() {
            @Override
            public int compare(Character o1, Character o2) {
                return priority[o1 - 97] - priority[o2 - 97];
            }
        };
        Character[] ans = new Character[T.length()];
        for (i = 0; i < T.length(); ++i) {
            ans[i] = new Character(T.charAt(i));
        }
        Arrays.sort(ans, letterComparator);
        StringBuilder stringBuilder = new StringBuilder();
        for (Character ch : ans) {
            stringBuilder.append(ch);
        }
        return stringBuilder.toString();
    }
}

解法二

统计字母出现次数，按照S中出现的次序填充，最后补上只在T中出现不再S中出现的字母。

import java.util.Arrays;
class Solution {
    public String customSortString(String S, String T) {
        // 统计字母出现次数
        int[] count = new int[26];
        int i;
        for (i = 0; i < T.length(); ++i) {
            ++count[T.charAt(i) - 97];
        }
        char ans[] = new char[T.length()];
        int len = 0;
        char ch;
        for (i = 0; i < S.length(); ++i) {
            ch = S.charAt(i);
            Arrays.fill(ans, len, len + count[ch - 97], ch);
            len += count[ch - 97];
            count[ch - 97] = 0;
        }
        for (i = 0; i < 26; ++i) {
            if (count[i] != 0) {
                Arrays.fill(ans, len, len + count[i], (char) (97 + i));
                len += count[i];
            }
        }
        return new String(ans);
    }
}