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.
代码
主要是用一个数组进行标记，标记其偶数位和基数位各个字母的个数，最后进行比较。
class Solution {
public int numSpecialEquivGroups(String[] A) {
HashSet<String> set = new HashSet<String>();
int len = A.length;
for(int i=0;i<len;i++){
int[] a = new int[52];
String x = A[i];
int len1 = x.length();
for(int j=0;j<len1;j++){
a[x.charAt(j)-'a'+26*(j%2)]++;
}
set.add(Arrays.toString(a));
}
return set.size();
}
}
