Problem,
You are given two strings order and s. All the words of order are unique and were sorted in some custom order previously.
Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.
Return any permutation of s that satisfies this property.
Solution,
교집합만 뽑아내서 하면 된다. 이것도 하도 옛날에 풀어서 다시 풀면 더 이쁘게 할 수 있을 것 같다.
class Solution {
public String customSortString(String order, String str) {
Map<Integer, Character> a = new HashMap<>();
Map<Character, Integer> o = new HashMap<>();
for(int i = 0; i < order.length(); ++i) {
o.put(order.charAt(i), i);
a.put(i, order.charAt(i));
}
StringBuilder result = new StringBuilder();
List<Integer> j = new ArrayList<>();
for(int i = 0; i < str.length(); ++i) {
char item = str.charAt(i);
if (o.containsKey(item)) j.add(o.get(item));
else result.append(item);
}
Collections.sort(j);
for(Integer item : j) {
result.append(a.get(item));
}
return result.toString();
}
}