LWC 73: 791. Custom Sort String
Problem:
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.
思路1：
调用官方排序接口，把先后顺序传入Comparator。
代码如下：
public String customSortString(String S, String T) {
Character[] cs = new Character[T.length()];
for (int i = 0; i < T.length(); ++i) {
cs[i] = T.charAt(i);
}
Arrays.sort(cs, new Comparator<Character>() {
@Override
public int compare(Character o1, Character o2) {
return S.indexOf(o1) - S.indexOf(o2);
}
});
StringBuilder sb = new StringBuilder();
for (char c : cs) sb.append(c);
return sb.toString();
}思路2：
先把T的每个字符和对应个数存入bucket数组，根据S的顺序扫一遍，最后再扫一遍剩余字符（非S中的字符）。
代码如下：
public String customSortString(String S, String T) {
int[] bucket = new int[26];
for (char c : T.toCharArray()) {
bucket[c - 'a']++;
}
StringBuilder sb = new StringBuilder();
for (char c : S.toCharArray()) {
for (int i = 0; i < bucket[c - 'a']; i++) {
sb.append(c);
}
bucket[c - 'a'] = 0;
}
for (int i = 0; i < 26; i++) {
for (int j = 0; j < bucket[i]; j++) {
sb.append((char) (i + 'a'));
}
}
return sb.toString();
}Python版本：
class Solution(object):
def customSortString(self, S, T):
"""
:type S: str
:type T: str
:rtype: str
"""
c, s = collections.Counter(T), set(S)
return ''.join(i * c[i] for i in S) + ''.join(i * c[i] for i in c if i not in s)
