題目描述 LeetCode 804

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on.

For convenience, the full table for the 26 letters of the English alphabet is given below:

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

Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, "cab" can be written as "-.-.-....-", (which is the concatenation "-.-." + "-..." + ".-"). We'll call such a concatenation, the transformation of a word.

Return the number of different transformations among all words we have.

Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation: 
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

There are 2 different transformations, "--...-." and "--...--.".

Note:

The length of words will be at most 100.
Each words[i] will have length in range [1, 12].
words[i] will only consist of lowercase letters.

中文描述

題目以摩爾斯電碼（Morse Code）為背景設(shè)題，摩爾斯電碼簡(jiǎn)單來(lái)說(shuō)是把26個(gè)小寫英文字母( 'a' ~ 'z' ) ，為它們建立了一一映射，從題目中所說(shuō)，'a' -> '.-' , 'b' -> '-...' , -- , 'z' -> '--..' .
本文的意思是根據(jù)每個(gè)字母的摩爾斯電碼，輸入為字符串?dāng)?shù)組，為每個(gè)字符串映射出相應(yīng)的電碼，并且要去掉重復(fù)的，最終返回這個(gè)字符串?dāng)?shù)組所映射成的不同的電碼個(gè)數(shù)。

解題思路

• 輸入字符串讀入 word 指針數(shù)組
• 摩爾斯每個(gè)字符的映射字符串讀入 wordsList 數(shù)組。
• 三重 for 循環(huán)，第一重從 word 中讀取字符串，第二重依次讀取從第一重循環(huán)中讀到的字符串的單個(gè)字符，第三重將從第二重讀到的單個(gè)字符映射為 wordsList 中該字符對(duì)應(yīng)的摩爾斯電碼，最終依次將 wordsList 中的映射依次讀入 key[i] 數(shù)組。
• a[] 用于計(jì)數(shù)，因題目要求去除冗余，所以，我先設(shè)置 a[i] 全為 1，從第一個(gè)依次往后都，如果后面和前面有重復(fù)的，就把 a[i] 設(shè)置為 0，最終 a[] 數(shù)組中為 1 的就是去掉重復(fù)的摩爾斯電碼。（簡(jiǎn)單來(lái)說(shuō)，可以把 a[] 數(shù)組理解為 key[] 的標(biāo)簽）

C語(yǔ)言實(shí)現(xiàn)

# include <stdio.h>
# include <string.h>

int uniqueMorseRepresentations(char** words, int wordsSize) 
{
    char wordsList[26][12] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
    int i = 0;
    int j = 0;
    int k = 0;
    char key[100][100];
    int size = 0;
    int a[100];
    
    // 把 words 中的每個(gè)字符串，找到字符串的每個(gè)字符在 wordsList 中的映射，把每個(gè)字符映射寫入 key[] 數(shù)組
    for ( i = 0; i < wordsSize; i ++)
    {
        size = 0;
        for ( j = 0; j < strlen( *(words + i) ); j ++)
        {
            //printf("%c ", *(*(words + i) + j));
            //printf("%s ", wordsList[ *(*(words + i) + j) - 'a' ]);
            
            for ( k = 0; k < strlen(wordsList[ *(*(words + i) + j) - 'a' ]); k ++)
            {
                key[i][size++] = wordsList[ *(*(words + i) + j) - 'a' ][k];
            }
        }
        key[i][size] = '\0';
        //printf("%d   %s\n\n", i, key[i]);
    }
        
    // a[] 數(shù)組初始化全為 1 ， 用于打標(biāo)簽
    for (i = 0; i < wordsSize; i ++)
    {
        a[i] = 1;
    }
    // 去除冗余，如果發(fā)現(xiàn)相應(yīng)有冗余，如 key[i] 和 key[j] 冗余，則將 a[j] 設(shè)為 0
    for (i = 0; i < wordsSize - 1; i ++)
    {
        if(a[i] == 1)
        {
            for (j = i + 1; j < wordsSize; j ++)
            {
                if( strcmp(key[i], key[j]) == 0)
                {
                    a[j] = 0;
                }
            }
        }
    }
        
    // 計(jì)數(shù)，最終在 a[] 數(shù)組中，為 1 的元素都是獨(dú)一無(wú)二的
    size = 0;
    for(i = 0; i < wordsSize; i ++)
    {
        if(a[i] == 1)
        {
            size ++ ;
        }
    }
    return size;
}

main()
{
      char *words[4] = {"gin", "zen", "gig", "msg"};

      printf("%d\n\n", uniqueMorseRepresentations(words, 4));
}

注意

• 這道題總體思考沒(méi)問(wèn)題，代碼也在 LeetCode執(zhí)行通過(guò)，但是在本地執(zhí)行會(huì)用問(wèn)題，估計(jì)是因?yàn)橹羔槹?，我看編譯器的報(bào)錯(cuò)，貌似要求 uniqueMorseRepresentations 這個(gè)函數(shù)，輸入為指針，輸出也要為指針，我真是 ***，當(dāng)然了，可能我才疏學(xué)淺，也可能是編譯器的原因，歡迎大牛指正。
• 對(duì)第一段更正， 經(jīng)過(guò)思考，終于發(fā)現(xiàn)錯(cuò)誤所在，真是基礎(chǔ)不牢，地動(dòng)山搖
  我的錯(cuò)誤是，最近寫 python 比較多，把 python 的輸出語(yǔ)法理所當(dāng)然地用到了 C 語(yǔ)言上
  python 中 print(666) 是沒(méi)問(wèn)題的。
  但是 在 C 語(yǔ)言中必須是 'printf("%d", 666)' ，不能直接 printf(666)
• 代碼已更正，本地可直接運(yùn)行。

吐槽

• C語(yǔ)言太不友好，我用C竟然能寫 50多行，在網(wǎng)上看別人用 C++ 才幾行，真是心哇涼哇涼的。 最近發(fā)現(xiàn)是 C 語(yǔ)言庫(kù)函數(shù)少的原因，就當(dāng)鍛煉編程能力了。
• 以后可能會(huì)轉(zhuǎn)戰(zhàn) C++ 啦，C 的指針真是博大精深，不過(guò)后話啦，據(jù)說(shuō) C 執(zhí)行效率最快，java 最慢。
• 網(wǎng)上有人建議，每做完一道題目去討論區(qū)看看高贊代碼，查漏補(bǔ)缺，不過(guò)現(xiàn)在真沒(méi)時(shí)間，等以后臨近找工作，再去看別人的代碼思考吧。