Code War每天一练第三天

Posted 2021-02-08 zch-boke

Take 2 strings s1 and s2 including only letters from ato z. Return a new sorted string, the longest possible, containing distinct letters,

• each taken only once - coming from s1 or s2.
• #Examples: ``` a = "xyaabbbccccdefww"   b = "xxxxyyyyabklmopq"
• longest(a, b) -> "abcdefklmopqwxy"

　　a = "abcdefghijklmnopqrstuvwxyz"

　　longest(a, a) -> "abcdefghijklmnopqrstuvwxyz" ```　

　　做题思路　

　　1、将字符串 s1和s2 链接成一个新字符串
　　2、去掉重复字符，返回新一个新无重复字符串
　　3、将字符串排序

 1 char * repeat(char *s)
 2 {
 3     //char *temp = s;
 4     int i = 0, j, k;
 5     int len = strlen(s);
 6 
 7     for (i = 0; i < len; i++)
 8     {
 9         for ( j = i + 1; j < len; j++)
10         {
11             if (s[i] == s[j])
12             {
13                 for ( k = j; k < len; k++)
14                 {
15                     s[k] = s[k + 1];
16                     //printf("%c  ", s[k]);
17                 }
18                 len--;
19                 j--; //注：如果j--的话那么下一次循环j++就跳过一次比较
20             }
21         }
22         
23     }
24     return s;
25  
26 }
27 
28 char* longest(char* s1, char* s2) {
29     // your code
30     //char *temp;
31     char *temp = (char *)malloc(strlen(s1)+strlen(s2)+1);
32     //strcat(s1, s2); //不能直接使用strcat否则可能产生写入冲突，需保证s1足够大
33     //将两字符串链接成一个字符串
34     strcpy(temp, s1);
35     strcat(temp, s2);
36     //strcat(s1, s2);
37     printf("%s
", s1);
38     temp = repeat(temp);
39     for (int i = 0; i < (int)strlen(temp); i++)
40     {
41         for (int j = i; j < (int)strlen(temp); j++)
42         {
43             if (temp[i] > temp[j])
44             {
45                 char tem = temp[i];
46                 temp[i] = temp[j];
47                 temp[j] = tem;
48             }
49         }
50     }
51     return temp;
52 
53 }

 这是一些大神的代码

 1 char* longest(char* s1, char* s2) {
 2   char letters [26] = {0}, *temp, *final = temp = (char *) calloc(sizeof(char), 26);;
 3   while(*s1) letters[*(s1++) - ‘a‘]++;
 4   while(*s2) letters[*(s2++) - ‘a‘]++;
 5   
 6   for(int i = 0; i < 26; i++)
 7     if(letters[i])
 8       *(temp++) = ‘a‘ + i;
 9       
10   return final;
11 }