c语言实现同一种字母只能出现 一次。 如:输入"deefd",结果为"def" 输入"ppppppp",结果为"p
时间: 2023-05-18 17:03:06 浏览: 126
这是一个经典的去重问题,可以使用哈希表来实现。具体的实现方法是,遍历字符串中的每个字符,将其作为哈希表的键,值可以随意设置,只要不是空即可。如果遍历到的字符已经在哈希表中出现过了,就跳过,否则将其加入结果字符串中。最后返回结果字符串即可。
以下是示例代码:
```c
#include <stdio.h>
#include <string.h>
char* removeDuplicates(char* s) {
int len = strlen(s);
int hash[26] = {0};
char* res = (char*)malloc(len + 1);
int idx = 0;
for (int i = 0; i < len; i++) {
int c = s[i] - 'a';
if (hash[c] == 0) {
res[idx++] = s[i];
hash[c] = 1;
}
}
res[idx] = '\0';
return res;
}
int main() {
char s1[] = "deefd";
char s2[] = "ppppppp";
printf("%s\n", removeDuplicates(s1)); // expect "def"
printf("%s\n", removeDuplicates(s2)); // expect "p"
return 0;
}
```
输出结果:
```
def
p
```
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