leetcode-893-Groups-of-Special-Equivalent-Strings

描述


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:

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

Example 2:

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

Example 3:

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

Example 4:

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

Note:

  • 1 <= A.length <= 1000
  • 1 <= A[i].length <= 20
  • All A[i] have the same length.
  • All A[i] consist of only lowercase letters.

分析


题目大意是给定一个包含字符串的集合,对于一些字符串,如果对于奇数位置的字符和偶数位置的字符任意调换顺序,得到的结果是一样的,那么这些字符串属于同一个组,要求根据输入的集合得出这个集合有几个组。

可以分别将字符串里的奇数位置的字符和偶数位置的字符进行排序,排序的结果放入一个哈希集合,最后这个哈希集合的大小就是组的数量。这个『排序』的过程可以用通过增加空间复杂度的方式实现,也可以用排序实现

解决方案1(Java)


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class Solution {
public int numSpecialEquivGroups(String[] A) {
Set<String> result = new HashSet();
for (String S: A) {
int[] count = new int[52];
for (int i = 0; i < S.length(); i++) {
count[S.charAt(i) - 'a' + 26*(i%2)]++;
}
result.add(Arrays.toString(count));
}
return result.size();
}
}

题目来源