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string xstring公司间的转换 XML ,常用类

Android开发指南 2024-09-16 阅读 5

给你一个字符串数组 words ,只返回可以使用在 美式键盘 同一行的字母打印出来的单词。键盘如下图所示。

美式键盘 中:

  • 第一行由字符 "qwertyuiop" 组成。
  • 第二行由字符 "asdfghjkl" 组成。
  • 第三行由字符 "zxcvbnm" 组成。
  • American keyboard

    示例 1:

    输入:words = ["Hello","Alaska","Dad","Peace"]
    输出:["Alaska","Dad"]
    

    示例 2:

    输入:words = ["omk"]
    输出:[]
    

    示例 3:

    输入:words = ["adsdf","sfd"]
    输出:["adsdf","sfd"]
  • #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    
    
    
    char** findWords(char** words, int wordsSize, int* returnSize)
    {
        char *ptr[] = {"qwertyuiop","asdfghjkl","zxcvbnm"};
        int pos1 = 0;
        int pos2 = 0;
        int count = 0;
        int start = 0;
        char **str = malloc(sizeof(char)*wordsSize);
    
        for(int i = 0;i < wordsSize;i++)//words行数
        {
            pos1 = 0;
            pos2 = 0;
            for(int j = 0;j < 3;j++)//ptr行数
            {
                while(words[i][pos1] != '\0' && ptr[j][pos2] != '\0')//将words的第一行第一个元素遍历ptr第一个每一个元素
                {
                    if(words[i][pos1] != ptr[j][pos2] && words[i][pos1] != ptr[j][pos2] - 32 )//不相等,ptr移向下一个元素
                    {
                        pos2++;
                    }
                    else if(words[i][pos1] == ptr[j][pos2] || words[i][pos1] == ptr[j][pos2] - 32)//相等,words移向下一个元素,ptr从第一个元素重新开始,计数+1
                    {
                        pos1++;
                        pos2 = 0;
                        count++;
                    }
                }
                if(count > 0 && count < strlen(words[i])-1)//一轮过后,如果计数存在并且,小于words的长度,说明words第一行元素不在同一行
                {
                    count = 0;
                    break;
                }
                if(count == strlen(words[i]))//
                {
                    str[start++] = words[i];
                    count = 0;
                    break;
                }
                pos1 = 0;
                pos2 = 0;
            }
        }
        *returnSize = start;
        return str;
    
    
    }
    int main()
    {
        char *words[] = {"Asd","zXc"};
        int wordsSize = sizeof(words)/sizeof(words[0]);
        int returnSize = 0;
        char ** p = findWords(words,wordsSize,&returnSize);
        for(int i = 0;i < returnSize;i++)
        {
            printf("%s  ",p[i]);
        }
        printf("\n");
        free(p);
        p = NULL;
        return 0;
    }

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