【算法千题案例】每日LeetCode打卡——95.唯一摩尔斯密码词

[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

输入: words = ["gin", "zen", "gig", "msg"]
输出: 2
解释: 
各单词翻译如下:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

共有 2 种不同翻译, "--...-." 和 "--...--.".

输入：words = ["a"]
输出：1

public class Solution
{
    public int UniqueMorseRepresentations(string[] words)
    {
        Dictionary<char, string> Dic = new Dictionary<char, string>()
        {{'a',".-"},{'b',"-..."},{'c',"-.-."},{'d',"-.."},{'e',"."},{'f',"..-."},{'g',"--."},{'h',"...."},
         {'i',".."},{'j',".---"},{'k',"-.-"},{'l',".-.."},{'m',"--"},{'n',"-."},{'o',"---"},{'p',".--."},
         {'q',"--.-"},{'r',".-."},{'s',"..."},{'t',"-"},{'u',"..-"},{'v',"...-"},{'w',".--"},{'x',"-..-"},
         {'y',"-.--"},{'z',"--.."}
        };//一个一个将26个字母所对应的摩斯密码输进去，key为字母，value为对应的密码。

        StringBuilder SB = new StringBuilder();//新建一个StringBuilder，经常改字符串用这个快一点，因为后面每个字母都要改字符串，所以这里用这个。
        Dictionary<String, int> Dic2 = new Dictionary<string, int>();//这个字典用来储存不相同的摩斯密码。
       

        foreach(string i in words)//遍历words里的所有单词。
        {
            for(int j = 0; j < i.Length; j++)//遍历单词的字母
            {
                //将每个字母所对应的摩斯密码连起来。
                SB.Append(Dic[  i[j]/* i[j]为字母，Dic[i[j]]代表这个字母在字典中的Value*/  ]  );
            }
          //↓遍历完一个单词后，如果Dic2中没有储存这条摩斯密码，则将其加入进去，value设置为1（这个随便，只要保证能知道这密码存在就行了）。
            if (!Dic2.ContainsKey(SB.ToString()))
            {
                Dic2.Add(SB.ToString(),1);
            }

         SB.Clear();//最后，将StringBuilder清空，因为遍历到下一串单词还用用上它。

        }

        return Dic2.Count;//最后只要返回Dic2的数量就行了，因为只加不重复的密码进去，所以它的数量就代表了不重复的密码。
    }
}
这里要感谢原文链接作者~
链接：https://leetcode-cn.com/problems/unique-morse-code-words/solution/bao-li-cha-zi-dian-cha-jiu-wan-shi-liao-by-huang-g/

通过
执行用时：88 ms，在所有 C# 提交中击败了25.50%的用户
内存消耗：35.4 MB，在所有 C# 提交中击败了29.90%的用户

class Solution {
    public int uniqueMorseRepresentations(String[] words) {
        String[] MORSE = new String[]{".-","-...","-.-.","-..",".","..-.","--.",
                         "....","..",".---","-.-",".-..","--","-.",
                         "---",".--.","--.-",".-.","...","-","..-",
                         "...-",".--","-..-","-.--","--.."};

        Set<String> seen = new HashSet();
        for (String word: words) {
            StringBuilder code = new StringBuilder();
            for (char c: word.toCharArray())
                code.append(MORSE[c - 'a']);
            seen.add(code.toString());
        }

        return seen.size();
    }
}

通过
执行用时：1 ms，在所有 Java  提交中击败了100.00%的用户
内存消耗：36.4 MB，在所有 Java 提交中击败了40.00%的用户

时间复杂度：O( n )
空间复杂度：O(1)