LeetCode 500. Keyboard Row

题目描述 LeetCode 500

Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

Note:

• You may use one character in the keyboard more than once.
• You may assume the input string will only contain letters of alphabet.

解题思路

• 依次取出 words 数字中的字符串，和 keyboard 中的每个字符串进行单个字符的比较。
• 例如，判断 Hello 是否在同一行，依次与 keyboard 中的字符串进行比较，如与 QWERTYUIOP 依次逐字符进行比较，设置 temp 计数变量，如 H 在 QWERTYUIOP 中则计数加 1，依次循环之后，如果 temp 等于 Hello 的长度，
  则 Hello 在同一行。

Code

# include
# include

int vv(char *word)
{
    int i, k, m;
    int flag = 0;
    int xx, yy, temp;
    char *keyboard[3] = {"QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"};

    xx = strlen(word);
    for (k = 0; k < 3; k ++)
    {
        yy = strlen(*(keyboard + k));
        temp = 0;
        for (m = 0; m < yy; m ++)
        {
            for (i = 0; i < xx; i++)
            {
                if ( (word[i] == *(*(keyboard + k) + m)) || (word[i] == (*(*(keyboard + k) + m) + 32)))
                {
                    temp += 1;                      
                }
            }

            if (temp == xx)
            {
                flag = 1;
                break;
            }
        }
    }

    if(flag == 1){
        return 1;
    }
    else{
        return 0;
    }
}

char** findWords(char** words, int wordsSize, int* returnSize) 
{
    int i ;
    static char *word[100];
    char** return_words;
    return_words = word;

    for (i = 0; i < wordsSize; i ++)
    {
        if(vv( *(words + i)))
        {
            // printf("%s\n", *(words + i));
            *(return_words + (*returnSize)) = *(words + i);
            (*returnSize) ++;
        }
    }
    
    return return_words;
}


int main()
{   
    int i ;
    int wordsSize = 4;
    int returnSize = 0;
    char *words[4] = {"Hello", "Alaska", "Dad", "Peace"};
    char** doo;
    
    doo = findWords(words, wordsSize, &returnSize);

    for (i = 0; i < returnSize; i ++){
        printf("%s\n", *(doo + i));
    }
}

运行结果