共计 1397 个字符，预计需要花费 4 分钟才能阅读完成。
题目
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:
Input: [“a”,”b”,”c”,”a”,”c”,”c”]
Output: 3
Explanation: 3 groups [“a”,”a”], [“b”], [“c”,”c”,”c”]
Example 2:
Input: [“aa”,”bb”,”ab”,”ba”]
Output: 4
Explanation: 4 groups [“aa”], [“bb”], [“ab”], [“ba”]
Example 3:
Input: [“abc”,”acb”,”bac”,”bca”,”cab”,”cba”]
Output: 3
Explanation: 3 groups [“abc”,”cba”], [“acb”,”bca”], [“bac”,”cab”]
Example 4:
Input: [“abcd”,”cdab”,”adcb”,”cbad”]
Output: 1
Explanation: 1 group [“abcd”,”cdab”,”adcb”,”cbad”]
Note:
1 <= A.length <= 1000
1 <= A[i].length <= 20
All A[i] have the same length.
All A[i] consist of only lowercase letters.
讲解
这道题我刚开始又没看懂，我以为是数组是一个字符串，对这个数组进行奇偶位的 swap。后来终于懂了，是对数组中的每个字符串进行奇偶位的 swap。
Java 代码
class Solution {
public int numSpecialEquivGroups(String[] A) {
Set<List> set = new HashSet<>();
for(String s:A){
char[] c = s.toCharArray();
List<Character> temp1 = new ArrayList<>();
if(c.length>2){
List<Character> temp2 = new ArrayList<>();
for(int i=0;i<c.length;i+=2){
temp1.add(c[i]);
}
Collections.sort(temp1);
for(int i=1;i<c.length;i+=2){
temp2.add(c[i]);
}
Collections.sort(temp2);
temp1.addAll(temp2);
}else{
for(int i=0;i<c.length;i++){
temp1.add(c[i]);
}
}
set.add(temp1);
}
return set.size();
}
}
