[Solution] 893. Groups of Special-Equivalent Strings

标签：ati   bool   cas   tostring   ==   ase   exp   letter   代码   

• Difficulty: Easy

Problem

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:

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"], ["abc", "bca"], ["bac", "cab"]

Example 4:

Input: ["abcd", "cdab", "adcb", "cbad"]
Output: 1
Explanation: 3 groups ["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.

Related Topics

String

Solution

统计每个字符串奇数位和偶数位上的字频，当两个字符串以此法统计的字频分布相同时，称这两个字符串为 special-equivalent。我的代码写得似乎麻烦了点，以后补充更简便的写法。

class CharCounter
{
    private SortedDictionary<char, int> counter;

    public CharCounter(string str)
    {
        counter = new SortedDictionary<char, int>();
        bool isOdd = false;

        foreach(char c in str)
        {
            if(isOdd)
            {
                if (counter.ContainsKey(char.ToUpper(c)))
                    ++counter[char.ToUpper(c)];
                else
                    counter.Add(char.ToUpper(c), 1);
            }
            else
            {
                if (counter.ContainsKey(c))
                    ++counter[c];
                else
                    counter.Add(c, 1);
            }
            isOdd = !isOdd;
        }
    }

    public override string ToString()
    {
        StringBuilder builder = new StringBuilder("{");
        foreach(var pair in counter)
        {
            builder.Append($"‘{pair.Key}‘:{pair.Value},");
        }
        builder.Remove(builder.Length - 1, 1);
        builder.Append("}");
        return builder.ToString();
    }

    public override bool Equals(object obj)
    {
        return ToString() == obj.ToString();
    }

    public override int GetHashCode()
    {
        return ToString().GetHashCode();
    }
}

public class Solution
{
    public int NumSpecialEquivGroups(string[] A)
    {
        return (from str in A select new CharCounter(str)).ToHashSet().Count;
    }
}

[Solution] 893. Groups of Special-Equivalent Strings

标签：ati   bool   cas   tostring   ==   ase   exp   letter   代码   

原文地址：https://www.cnblogs.com/Downstream-1998/p/10404047.html