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 jwith 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 <= 10001 <= A[i].length <= 20- All
A[i]have the same length. - All
A[i]consist of only lowercase letters.
这道题定义了一种特殊相等的关系，就是说对于一个字符串，假如其偶数位字符之间可以互相交换，且其奇数位字符之间可以互相交换，交换后若能跟另一个字符串相等，则这两个字符串是特殊相等的关系。现在给了我们一个字符串数组，将所有特殊相等的字符串放到一个群组中，问最终能有几个不同的群组。最开始的时候博主没仔细审题，以为是随意交换字母，就直接对每个单词进行排序，然后扔到一个 HashSet 中就行了。后来发现只能是奇偶位上互相交换，于是只能现先将奇偶位上的字母分别抽离出来，然后再进行分别排序，之后再合并起来组成一个新的字符串，再丢到 HashSet 中即可，利用 HashSet 的自动去重复功能，这样最终留下来的就是不同的群组了，参见代码如下：
class Solution {
public:
int numSpecialEquivGroups(vector<string>& A) {
unordered_set<string> st;
for (string word : A) {
string even, odd;
for (int i = 0; i < word.size(); ++i) {
if (i % 2 == 0) even += word[i];
else odd += word[i];
}
sort(even.begin(), even.end());
sort(odd.begin(), odd.end());
st.insert(even + odd);
}
return st.size();
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/893
参考资料：
https://leetcode.com/problems/groups-of-special-equivalent-strings/
[LeetCode All in One 题目讲解汇总(持续更新中...)](https://www.cnblogs.com/grandyang/p/4606334.html)