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《LeetCode刷题(Easy Rank):893. Groups of Special-Equivalent Strings》

 2月前  •   LeetCode  •     •   11  •   0

Question:

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:

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

Example 2:

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

Example 3:

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

Example 4:

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

Solution:

如果两个数字分别进行奇偶位上的交换后能相等,那么这两个数字是同一组的。判断给定的字符数组中有多少组。

JAVAclass Solution {
    public int numSpecialEquivGroups(String[] A) {
       Set<String> set = new HashSet<>();
        for(String a : A){
            int n = a.length();
            char[] odd = new char[(n + 1)/2];
            char[] even = new char[(n + 1)/2];
            int to = 0, te = 0;
            for(int i = 0; i < a.length(); i++){
                if(i % 2 == 0) even[te++] = a.charAt(i);
                else odd[to++] = a.charAt(i);
            }
            Arrays.sort(odd);
            Arrays.sort(even);
            set.add(String.valueOf(odd)+String.valueOf(even));
        }
        return set.size();
    }
}

 

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