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## 时光荏苒，我们一直都在 ### 《LeetCode刷题（Easy Rank）：893. Groups of Special-Equivalent Strings》

3月前  •   •   •   33  •

### Question:

You are given an array A of strings.Two strings S and T are special-equivalent if after any number of moves, S == T.A move consists of choosing two indices i and j with i % 2 == j % 2, and swapping S[i] with S[j].

Now, a group of special-equivalent strings from A is a non-empty subset S of A such that any string not in S is not special-equivalent with any string in S.Return the number of groups of special-equivalent strings from A.

### Example 1:

示例Input: ["a","b","c","a","c","c"]
Output: 3
Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]

### Example 2:

示例Input: ["aa","bb","ab","ba"]
Output: 4
Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]

### Example 3:

示例Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]

### Example 4:

示例Input: ["abcd","cdab","adcb","cbad"]
Output: 1
Explanation: 1 group ["abcd","cdab","adcb","cbad"]

### Solution:

JAVAclass Solution {
public int numSpecialEquivGroups(String[] A) {
Set<String> set = new HashSet<>();
for(String a : A){
int n = a.length();
char[] odd = new char[(n + 1)/2];
char[] even = new char[(n + 1)/2];
int to = 0, te = 0;
for(int i = 0; i < a.length(); i++){
if(i % 2 == 0) even[te++] = a.charAt(i);
else odd[to++] = a.charAt(i);
}
Arrays.sort(odd);
Arrays.sort(even);
}  