[LeetCode]893. Groups of Special-Equivalent Strings
原題
You are given an arrayA
of strings.
Two stringsS
and T
are special-equivalent
if after any number of moves
, S == T.
Amove
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 fromA
.
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 A[i] A[i]
題解
明確題意
A。
然後，定義一個Special-Equivalent概念，如果有兩個字串S和T，將S串中的奇數位置相互移動若干次或不移動，偶數位置相互移動若干次或不移動之後，可以使兩個字串S和T相等，那麼稱為S和T是Special Equivalent。
思路
奇數位和偶數位是兩個分類，但是是一個緯度，如果解決了奇數位，那麼就解決了偶數位。
所以，先考慮偶數位。SE（even for string S）和TE（odd for string T）。判斷SE和TE能否通過調整各自內部字元的位置達到一致的問題，其實就是判斷SE和TE的組成元素是否一致，即排序後的字串是否一樣。同理，判斷奇數位的組成元素（字元）。
程式碼
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();
}
};
複雜度
M 代表陣列A的長度
N 代表A陣列中最長的字串的長度
時間複雜度：O(M*N*lgN)
原題：ofollow,noindex">https://leetcode.com/problems/groups-of-special-equivalent-strings/
文章來源：胡小旭 => [LeetCode]893. Groups of Special-Equivalent Strings