You are given an array A
of strings.html
Two strings S
and T
are special-equivalent if after any number of moves, S == T.git
A move consists of choosing two indices i
and j
with i % 2 == j % 2
, and swapping S[i]
with S[j]
.github
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.數組
Return the number of groups of special-equivalent strings from A
.app
Example 1:ui
Input: ["a","b","c","a","c","c"] Output: 3 Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]
Example 2:code
Input: ["aa","bb","ab","ba"] Output: 4 Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]
Example 3:htm
Input: ["abc","acb","bac","bca","cab","cba"] Output: 3 Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]
Example 4:blog
Input: ["abcd","cdab","adcb","cbad"] Output: 1 Explanation: 1 group ["abcd","cdab","adcb","cbad"]
Note:排序
1 <= A.length <= 1000
1 <= A[i].length <= 20
A[i]
have the same length.A[i]
consist of only lowercase letters.
這道題定義了一種特殊相等的關係,就是說對於一個字符串,假如其偶數位字符之間能夠互相交換,且其奇數位字符之間能夠互相交換,交換後若能跟另外一個字符串相等,則這兩個字符串是特殊相等的關係。如今給了咱們一個字符串數組,將全部特殊相等的字符串放到一個羣組中,問最終能有幾個不一樣的羣組。最開始的時候博主沒仔細審題,覺得是隨意交換字母,就直接對每一個單詞進行排序,而後扔到一個 HashSet 中就好了。後來發現只能是奇偶位上互相交換,因而只能現先將奇偶位上的字母分別抽離出來,而後再進行分別排序,以後再合併起來組成一個新的字符串,再丟到 HashSet 中便可,利用 HashSet 的自動去重複功能,這樣最終留下來的就是不一樣的羣組了,參見代碼以下:
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
參考資料:
https://leetcode.com/problems/groups-of-special-equivalent-strings/