原题

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.

题解

明确题意

给定一个字符串向量vector A。
然后，定义一个Special-Equivalent概念，如果有两个字符串S和T，将S串中的奇数位置相互移动若干次或不移动，偶数位置相互移动若干次或不移动之后，可以使两个字符串S和T相等，那么称为S和T是Special Equivalent。

思路

奇数位和偶数位是两个分类，但是是一个纬度，如果解决了奇数位，那么就解决了偶数位。
所以，先考虑偶数位。SE（even for string S）和TE（odd for string T）。判断SE和TE能否通过调整各自内部字符的位置达到一致的问题，其实就是判断SE和TE的组成元素是否一致，即排序后的字符串是否一样。同理，判断奇数位的组成元素（字符）。

代码

[code lang=”cpp”]
class Solution {
public:
int numSpecialEquivGroups(vector<string>& A) {
map<string, int> a;
string es, os;
for (auto str : A) {
es = os = "";
for (int i = 0; i < str.length(); ++i) {
if (i % 2 == 0) es += str[i];
if (i % 2 == 1) os += str[i];
}
sort(es.begin(), es.end());
sort(os.begin(), os.end());
a[es + os] += 1;
}
return a.size();
}
};
[/code]