[LeetCode]893. Groups of Special-Equivalent Strings

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 .

Input: ["a","b","c","a","c","c"]
Output: 3
Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]

Input: ["aa","bb","ab","ba"]
Output: 4
Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]

Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]

Input: ["abcd","cdab","adcb","cbad"]
Output: 1
Explanation: 1 group ["abcd","cdab","adcb","cbad"]

1 <= A.length <= 1000
1 <= A[i].length <= 20
A[i]
A[i]

题解

明确题意

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

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();
    }
};

复杂度

M 代表数组A的长度

N 代表A数组中最长的字符串的长度

时间复杂度：O(M*N*lgN)

原题： https://leetcode.com/problems/groups-of-special-equivalent-strings/

文章来源： 胡小旭 =>  [LeetCode]893. Groups of Special-Equivalent Strings